Concise Mathematics Selina Solutions Class 10 ICSE Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode)

This chapter provides comprehensive ICSE Class 10 Mean Median Mode Solutions from the Selina Concise Mathematics textbook. In statistics, we often need a single value to represent a whole group of data, and that’s where measures of central tendency come in. You will learn to calculate the three main measures: the mean (the average), the median (the middle value), and the mode (the most frequent value). For Class 10, we go deeper into finding these for both ungrouped and grouped data, using methods like the direct method, assumed mean method, and step-deviation method. We will also explore quartiles and how to find the median graphically using cumulative frequency curves, or ogives, which is a very important skill for your board exams.

If you are stuck on a specific question about calculating the mean of a frequency distribution or finding the median from an ogive, you have come to the right place. This page contains detailed, step-by-step solutions for all 66 questions from Exercise 24(A), 24(B), 24(C), and the Test Yourself section. Each solution is worked out using the exact same methods and formulae that the ICSE board expects you to use in your examinations. Here you will find clear, accurate, and easy-to-follow solutions to help you master every concept in Measures of Central Tendency.

Exercise 24(A)

Question 1(a)

The mean of 5 variables is 20. If four of them are 15, 25, 10 and 12; the fifth variable is :

• (a) 38
• (b) 60
• (c) 48
• (d) 28

Assume the fifth variable is denoted by x.

Using the formula for mean:

Mean = \dfrac{\text{Sum of observations}}{\text{No. of observations}}

Given that the mean is 20, we can write:

Multiply both sides by 5 to clear the fraction:

This simplifies to:

Subtract 62 from both sides to solve for x:

Hence, Option 1 is the correct option.

Question 1(b)

The mean height of 5 students is 140 cm. If height of one student is 156 cm, the mean height of remaining students is :

• (a) 74 cm
• (b) 136 cm
• (c) 16 cm
• (d) 80 cm

To find the mean, we use the formula:

Mean = \dfrac{\text{Sum of observations}}{\text{No. of observations}}

Here, the mean height of 5 students is given as 140 cm.

∴ \text{Mean height} = \dfrac{\text{Total height}}{\text{No. of students}}

⇒ 140 = \dfrac{\text{Total height}}{5}

This implies:

⇒ \text{Total height} = 700 \text{ cm}.

We know one student’s height is 156 cm. Therefore, the total height of the remaining 4 students is:

700 - 156 = 544 \text{ cm}.

Now, to find the mean height of these 4 students:

Mean height of remaining students = \dfrac{\text{Total height of 4 students}}{4} = \dfrac{544}{4} = 136 \text{ cm}.

Hence, Option 2 is the correct option.

Question 1(c)

The mean age of eight boys is 16 years. If two more boys, with ages 18 years and 14 years, join them, the resulting mean age is :

• (a) 18 years
• (b) 14 years
• (c) 16 years
• (d) 32 years

To find the mean, we use the formula:

Mean = \dfrac{\text{Sum of observations}}{\text{No. of observations}}

We know that the mean age of the eight boys is 16 years. This can be expressed as:

Now, two additional boys join the group, aged 18 years and 14 years respectively.

Adding these ages to the previous total, we get:

Total age now = 128 + 18 + 14 = 160.

The new mean age is given by:

New mean age = \dfrac{\text{New total age}}{\text{No. of boys}} = \dfrac{160}{10} = 16 years.

Hence, Option 3 is the correct option.

Question 1(d)

The mean value of 15 numbers is 20. If one of these numbers is wrongly taken as 45 instead of 15, the correct mean is :

• (a) 30
• (b) 15
• (c) 20
• (d) 18

To find the mean, we use the formula:

Mean = \dfrac{\text{Sum of observations}}{\text{No. of observations}}

Here, it is given that the mean of 15 numbers is 20.

\Rightarrow \text{Sum of observations} = 20 \times 15 = 300.

It is mentioned that one of the numbers was incorrectly recorded as 45 instead of 15.

Thus, the correct sum of observations should be:

Correct sum = 300 - 45 + 15 = 270.

Now, calculate the correct mean:

Correct mean = \dfrac{\text{Correct sum of observations}}{\text{No. of observations}}

\Rightarrow \text{Correct mean} = \dfrac{270}{15} = 18.

Hence, Option 4 is the correct option.

Question 1(e)

The mean of the given frequency distribution is :

• (a) \dfrac{30}{15} + \dfrac{20}{5}
• (b) \dfrac{30}{15} - \dfrac{20}{5}
• (c) \dfrac{550}{20} = 27.5
• (d) \dfrac{50}{20} = 2.5

To find the mean of the given frequency distribution, we first calculate the product of each value of x and its corresponding frequency f, then sum these products. The table below shows these calculations:

The formula for the mean of a frequency distribution is:

Mean = \dfrac{Σfx}{Σf}

Substituting the values, we have:

Mean = \dfrac{550}{20} = 27.5

Option 3 is the correct option.

Question 2

Marks obtained (in mathematics) by 9 students are given below :

60, 67, 52, 76, 50, 51, 74, 45 and 56.

(a) Find the arithmetic mean.

(b) If marks of each student be increased by 4; what will be the new value of arithmetic mean?

(a) To find the arithmetic mean, we first calculate the total of all the marks: 60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56 = 531.

Using the formula for mean:

Mean = \dfrac{\text{Sum of observations}}{\text{Number of observations}}

= \dfrac{531}{9} = 59.

Hence, mean = 59.

(b) Notice that if we increase each student’s marks by the same amount, the mean will also increase by that amount. Here, each mark is increased by 4.

∴ The mean will increase by 4.

New mean = 59 + 4 = 63.

Hence, new mean = 63.

Question 3

(a) Find the mean of 7, 11, 6, 5 and 6.

(b) If each number given in (a) is diminished by 2; find the new value of mean.

(a) To find the mean, first calculate the total of the numbers: 7 + 11 + 6 + 5 + 6 = 35.

Using the formula for mean:

Mean = \dfrac{\text{Sum of observations}}{\text{Number of observations}}

= \dfrac{35}{5} = 7.

Thus, the mean is 7.

(b) An important concept to remember is that if each number in a dataset is changed by the same amount, the mean will also change by that same amount.

∴ When each number is reduced by 2, the mean also reduces by 2.

New mean = 7 – 2 = 5.

Therefore, the new mean is 5.

Question 4

If the mean of 6, 4, 7, a and 10 is 8, find the value of ‘a’.

To find the value of ‘a’, we start by calculating the sum of the numbers: 6, 4, 7, a, and 10. This gives us a total of a + 27.

According to the formula for the mean:

Mean = \dfrac{\text{Sum of observations}}{\text{Number of observations}}

We know the mean is 8 and there are 5 numbers. Thus, we have:

⇒ 8 = \dfrac{a + 27}{5}

Multiplying both sides by 5 to clear the fraction, we get:

⇒ 40 = a + 27

Subtracting 27 from both sides, we find:

⇒ a = 13.

Hence, the value of a = 13.

Question 5

The mean of the number 6, y, 7, x and 14 is 8. Express y in terms of x.

To find the mean, we first calculate the sum of the numbers: 6, y, 7, x, and 14. This gives us a total of x + y + 27.

Using the formula for mean:

Mean = \dfrac{\text{Sum of observation}}{\text{No. of observation}}

Given the mean is 8, we have:

⇒ 8 = \dfrac{x + y + 27}{5}

To eliminate the fraction, multiply both sides by 5:

⇒ 40 = x + y + 27

Subtract 27 from both sides to isolate the terms with variables:

⇒ x + y = 40 – 27

⇒ x + y = 13

Now, solve for y by subtracting x from both sides:

⇒ y = 13 – x.

Hence, y = 13 – x.

Question 6

The ages of 40 students are given in the following table :

Find the arithmetic mean.

To determine the arithmetic mean of students’ ages, we first organize the given data into a table that includes the product of each age and its corresponding frequency. Here’s how it looks:

Notice that the total of the frequencies, Σf, is 40, which represents the number of students. The sum of the products of ages and frequencies, Σfx, is 614.

The formula for calculating the mean is:

Mean = \dfrac{Σfx}{n}

Substituting the values, we get:

Mean = \dfrac{614}{40} = 15.35

Hence, arithmetic mean = 15.35

Question 7

If 69.5 is the mean of 72, 70, x, 62, 50, 71, 90, 64, 58 and 82 : find the value of x.

To find the value of x, we first calculate the total sum of the numbers given: 72, 70, x, 62, 50, 71, 90, 64, 58, and 82. Adding these together, we get 619 + x.

According to the formula for the mean:

Mean = \dfrac{\text{Sum of observations}}{\text{Number of observations}}

Given that the mean is 69.5, we can set up the equation:

⇒ 69.5 = \dfrac{619 + x}{10}

Multiplying both sides by 10 to clear the fraction gives:

⇒ 695 = x + 619

Subtracting 619 from both sides, we find:

⇒ x = 695 – 619

⇒ x = 76.

Hence, x = 76.

Question 8

The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of ‘f’.

Let’s organize the data in a table to calculate the necessary values:

Notice that the total number of plants, denoted as n, is given by Σf = 28 + f.

The formula for the mean is:

Substituting the known values, we have:

Multiplying both sides by 28 + f to clear the fraction:

Expanding the left side gives:

Rearranging terms to isolate f:

Simplifying further, we find:

Finally, solving for f:

Thus, the value of f is 12.

Question 9

From the data, given below, calculate the mean wage, correct to the nearest rupee.

(i) If the number of workers in each category is doubled, what would be the new mean wage ?

(ii) If the wages per day in each category are increased by 60%; what is the new mean wage ?

(iii) If the number of workers in each category is doubled and the wages per day per worker are reduced by 40%; what would be the new mean wage ?

To find the mean wage, use the formula:

Mean = \dfrac{Σfx}{Σf} = \dfrac{3360}{42} = 80.

(i) The original mean is calculated as \dfrac{Σfx}{Σf}.

If the number of workers in each category doubles, the new mean becomes:

New mean = \dfrac{2Σfx}{2Σf} = \dfrac{Σfx}{Σf}, which is the same as the original mean.

∴ If the number of workers in each category is doubled, the mean wage remains unchanged.

Hence, mean = 80.

(ii) When wages per day increase by 60%, the mean wage also increases by 60%.

Calculate the new mean as follows:

New mean = 80 + \dfrac{60}{100} \times 80

= 80 + 48 = 128.

Hence, new mean = 128.

(iii) Doubling the number of workers does not affect the mean.

If wages decrease by 40%, the mean decreases by 40% as well.

New mean = 80 – \dfrac{40}{100} \times 80

= 80 – 32

= 48.

Hence, new mean = 48.

Question 10

The contents of 100 match boxes were checked to determine the number of matches they contained.

(i) Calculate, correct to one decimal place, the mean number of matches per box.

(ii) Determine, how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches ?

(i)

To find the mean number of matches per box, use the formula:

Mean = \dfrac{Σfx}{Σf}, where Σfx is the sum of the product of matches and boxes, and Σf is the total number of boxes.

∴ Mean = \dfrac{3813}{100} = 38.13 ≈ 38.1

Thus, the mean is 38.1.

(ii) To determine how many additional matches are needed to achieve a mean of 39 matches per box, consider the formula:

Mean = \dfrac{\text{No. of matches}}{\text{No. of boxes}}

Given the desired mean is 39:

⇒ 39 = \dfrac{\text{No. of matches}}{100}

⇒ No. of matches = 3900

The additional matches required = 3900 – 3813 = 87.

Therefore, 87 extra matches must be added.

Question 11

If the mean of the following distribution is 3, find the value of p.

To solve this, let’s first complete the frequency table with the product of each value of x and its corresponding frequency f:

The mean of a distribution is given by the formula:

Mean = \dfrac{Σfx}{Σf}

Given that the mean is 3, we substitute the known values into the formula:

⇒ 3 = \dfrac{6p + 87}{33}

Multiplying both sides by 33 to clear the fraction, we have:

⇒ 99 = 6p + 87

Subtracting 87 from both sides gives:

⇒ 12 = 6p

To find p, divide both sides by 6:

⇒ p = \dfrac{12}{6}

⇒ p = 2.

Hence, p = 2.

Question 12

In the following table, Σf = 200 and mean = 73. Find the missing frequencies f~1 and f~2.

Let’s organize the data into a table with the additional column for the product of x and f:

We know that the sum of the frequencies, \Sigma f, is 200.

This simplifies to:

From this equation, we can express f_1 in terms of f_2:

The mean is given as 73, which implies:

Substituting the known values:

This equation becomes:

Simplifying further:

Dividing both sides by 50 gives:

Substitute the expression for f_1 from equation (1):

This simplifies to:

Solving for f_2:

Substituting f_2 = 38 back into equation (1):

Hence, f_1 = 76 and f_2 = 38.

Question 13

Find the arithmetic mean (correct to nearest whole number) by using step deviation method.

Consider the assumed mean (A) to be 25.

Using the formula for the mean:

Thus, the mean is 22.

Question 14

Find the mean (correct to one place of decimal) by using short-cut method.

Assume the mean (A) to be 45.

Using the shortcut formula for mean:

Calculate:

Thus, the mean is:

Hence, mean = 44.6

Exercise 24(B)

Question 1(a)

The mean of given observations is :

• (a) 41
• (b) 27
• (c) 71
• (d) 91

To find the mean, we apply the formula:

Mean = \dfrac{Σfx}{Σf} = \dfrac{205}{5} = 41.

Hence, Option 1 is the correct option.

Question 1(b)

For data given in the adjoining table, the mean is :

• (a) 17
• (b) 27
• (c) 25
• (d) 60

To find the mean from the given table, we need to first determine the class mark x for each class interval. The class mark is the midpoint of each class interval. Here, the class size i is 10.

The mean is calculated using the formula:

Substituting the given values into the formula, we have:

Thus, the mean of the data is 17. Hence, Option 1 is the correct option.

Question 1(c)

The mean of observations, given in the adjoining table, is 20, the value of a is :

• (a) 20
• (b) 30
• (c) 10
• (d) 25

To find the mean, we use the formula:

Mean = \dfrac{\Sigma fx}{\Sigma f}

Here’s the table with calculations:

Substituting these into the mean formula:

Multiplying both sides by (20 + a) gives:

Simplifying, we get:

Notice that this equation is an identity, meaning it holds true for any value of a. Thus, the problem as stated cannot be solved to find a specific value for a, indicating an error in the question setup.

Question 1(d)

If the mean of the data given in adjoining table is 20, the relation between x~1 and x~2 is :

• (a) x~1 + x~2 = 30
• (b) x~1 – x~2 = 15
• (c) x~1 – x~2 = 0
• (d) x~1 + x~2 = 20

Let’s analyze the data provided in the table:

The mean of the dataset is calculated using the formula:

Mean = \dfrac{Σfx}{Σf}

Given that the mean is 20, substitute the known values into the equation:

Notice that the calculations show that x_1 must be equal to x_2. This gives us the relationship between x_1 and x_2.

Hence, Option 3 is the correct option.

Question 1(e)

The mean of data, represented by given diagram is :

1. 53
2. 47
3. 42
4. 51

Let’s organize the information from the graph into a tabular format:

To find the mean, use the formula:

Mean = \dfrac{Σfx}{Σf} = \dfrac{2350}{50} = 47.

Hence, Option 2 is the correct option.

Question 2

The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.

To determine the arithmetic mean of the ages, let’s first calculate the midpoints for each age group. These midpoints will represent the average age for each interval.

The formula for the arithmetic mean is:

Mean = \dfrac{Σfx}{Σf}

Substituting the values, we have:

Mean = \dfrac{1074}{50} = 21.48

Hence, mean = 21.48

Question 3

The following are the marks obtained by 70 boys in a class test.

Calculate the mean by :

(i) Short-cut method

(ii) Step-deviation method

(i) Let’s assume the mean (A) is 65.

The total number of boys, n = Σf = 70.

Using the formula for the assumed mean method:

Mean = A + \dfrac{Σfd}{n} = 65 + \dfrac{-270}{70}

= 65 – 3.86 = 61.14

So, the mean is 61.14.

(ii) For the step-deviation method, we’ll use the same assumed mean (A) = 65 and a class interval (i) of 10.

Here, n = Σf = 70.

Applying the step-deviation formula:

Mean = A + \dfrac{Σft}{n} \times i

= 65 + \dfrac{-27}{70} \times 10

= 65 – \dfrac{27}{7}

= 65 – 3.86

= 61.14

This gives us the same mean of 61.14.

Question 4

Find mean by ‘step-deviation method’ :

To find the mean using the step-deviation method, let’s take the assumed mean (A) as 87.5 and the class interval width (i) as 7.

Here, the total frequency (n) is 160.

Using the formula for the mean:

Mean = A + \dfrac{Σft}{n} \times i

Substitute the values:

= 87.5 + \dfrac{29}{160} \times 7

= 87.5 + \dfrac{203}{160}

= 87.5 + 1.3

= 88.8

Therefore, the mean is 88.8.

Question 5

The mean of following frequency distribution is 21\dfrac{1}{7}. Find the value of ‘f’.

To find the value of ‘f’, let’s first calculate the class marks (x) for each class interval, which are the midpoints of the intervals:

The mean of the distribution is given by the formula:

Mean = \dfrac{Σfx}{Σf}

We know the mean is 21\dfrac{1}{7}, thus:

\Rightarrow 21\dfrac{1}{7} = \dfrac{1235 + 35f}{63 + f}
\Rightarrow \dfrac{148}{7} = \dfrac{1235 + 35f}{63 + f}
\Rightarrow 148(63 + f) = 7(1235 + 35f)
\Rightarrow 9324 + 148f = 8645 + 245f

Now, let’s solve for ‘f’:

\Rightarrow 245f - 148f = 9324 - 8645
\Rightarrow 97f = 679
\Rightarrow f = \dfrac{679}{97} = 7.

Therefore, the value of ‘f’ is 7.

Question 6

Using the information given in the adjoining histogram; calculate the mean.

To determine the mean from the histogram data, we first organize the information into a table:

The formula for calculating the mean is:

Mean = \dfrac{Σfx}{Σf}

Substitute the values from our table:

Mean = \dfrac{2850}{75} = 38.

Thus, the mean is 38.

Question 7

If the mean of the following observations is 54, find the value of p.

To determine the value of p, we start by calculating the class mark for each class interval using the formula:

Class mark = \dfrac{\text{Upper limit + Lower limit}}{2}.

Let’s organize the data in a table:

The mean is given by the formula:

Mean = \dfrac{Σfx}{Σf}

Given that the mean is 54, we set up the equation:

⇒ 54 = \dfrac{2370 + 30p}{39 + p}

Multiplying both sides by (39 + p), we have:

⇒ 54(39 + p) = 2370 + 30p

Expanding the left-hand side, we get:

⇒ 2106 + 54p = 2370 + 30p

Rearranging gives:

⇒ 54p – 30p = 2370 – 2106

⇒ 24p = 264

Solving for p:

⇒ p = \dfrac{264}{24}

⇒ p = 11.

Thus, the value of p is 11.

Question 8

The mean of the following distribution is 62.8 and the sum of all the frequencies is 50. Find the missing frequencies f~1 and f~2.

To determine the missing frequencies f_1 and f_2, we start by calculating the class mark for each class interval using the formula:

Class mark = \dfrac{\text{Upper limit} + \text{Lower limit}}{2}

Here is the table with class marks and their corresponding contributions to the total sum:

We know the total sum of frequencies is 50, so:

∴ f_1 + f_2 + 30 = 50

⇒ f_1 + f_2 = 20

⇒ f_1 = 20 - f_2 ……..(1)

The mean of the distribution is given by:

Mean = \dfrac{Σfx}{Σf}

∴ 62.8 = \dfrac{2060 + 30f_1 + 70f_2}{50}

⇒ 2060 + 30f_1 + 70f_2 = 3140

⇒ 30f_1 + 70f_2 = 1080

Substituting the expression for f_1 from equation (1) into the above equation:

⇒ 30(20 - f_2) + 70f_2 = 1080

⇒ 600 - 30f_2 + 70f_2 = 1080

⇒ 40f_2 = 480

⇒ f_2 = \dfrac{480}{40} = 12

Using the value of f_2, we find f_1:

⇒ f_1 = 20 - f_2 = 20 - 12 = 8

Hence, f_1 = 8 and f_2 = 12.

Question 9

Calculate the mean of the distribution, given below, using the short cut method :

To find the mean using the shortcut method, we first need to adjust the given distribution since it is discontinuous. The adjustment factor is calculated as follows:

Adjustment factor = \dfrac{\text{Lower limit of one class} - \text{Upper limit of previous class}}{2}

= \dfrac{21 - 20}{2} = \dfrac{1}{2}

= 0.5

Now, we adjust the class limits by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit.

Assume the mean (A) as 45.5.

The total frequency, n = \Sigma f = 50.

Now, calculate the mean using:

Mean = A + \dfrac{\Sigma fd}{n}

= 45.5 + \dfrac{70}{50}

= 45.5 + 1.4

= 46.9

Hence, mean = 46.9

Exercise 24(C)

Question 1(a)

The median of 18, 29, 15, 14 and 21 is :

• (a) 15
• (b) 18
• (c) 29
• (d) 21

First, sort the numbers in increasing order: 14, 15, 18, 21, 29.

Since there are 5 numbers, which is an odd count, the median is the middle term.

∴ Median = \dfrac{n + 1}{2} th term, where n = 5.

Substituting the values, we calculate:

Median = \dfrac{5 + 1}{2} = \dfrac{6}{2} = 3rd term.

The 3rd term in the sequence is 18.

Hence, Option 2 is the correct option.

Question 1(b)

The median of 3, 8, 11, 2, 16, 4, 0 and 6 is :

• (a) 6
• (b) 9
• (c) 5
• (d) 8

First, let’s arrange the given numbers in increasing order: 0, 2, 3, 4, 6, 8, 11, 16.

Notice that there are 8 numbers in total, which is an even count.

∴ Median is calculated using the formula for even number of terms:

Substituting the values, we have:

Hence, Option 3 is the correct option.

Question 1(c)

Numbers 5, 15, 20, x, 28, 30, 35 are in ascending order and have median = 23; then the value of x is :

• (a) 24
• (b) 29
• (c) 17.5
• (d) 23

We have the sequence of numbers: 5, 15, 20, x, 28, 30, 35, which are arranged in increasing order.

Since there are 7 numbers in total, which is an odd count, the median is found at the position given by \dfrac{n + 1}{2}, where n is the number of terms.

Here, n = 7, so:

Hence, Option 4 is the correct option.

Question 1(d)

For numbers 10, 20, 30, 40, 50, 60, 70 and 80; the inter-quartile range is :

• (a) 20 + 60
• (b) 60 – 30
• (c) 60 – 20
• (d) 50 – 10

Arranging the numbers in increasing order gives us: 10, 20, 30, 40, 50, 60, 70, and 80.

Here, the total count of numbers, n, is 8, which is an even number.

∴ The lower quartile (Q1) is determined by \dfrac{n}{4} = \dfrac{8}{4} = 2. This corresponds to the 2nd term, which is 20.

Similarly, the upper quartile (Q3) is found using \dfrac{3n}{4} = \dfrac{3 \times 8}{4} = 3 \times 2 = 6. This matches the 6th term, which is 60.

Thus, the inter-quartile range is calculated as the difference between the upper quartile and the lower quartile: 60 – 20.

Hence, Option 3 is the correct option.

Question 1(e)

From the given diagram, the modal class is :

1. 30 – 40
2. 40 – 50
3. 50 – 60
4. 60 – 70

Observing the diagram, it is evident that the class interval 50-60 exhibits the greatest frequency among all the intervals.

Hence, Option 3 is the correct option.

Question 2

A student got the following marks in 9 questions of a question paper.
3, 5, 7, 3, 8, 0, 1, 4 and 6. Find the median of these marks.

First, let’s put the marks in order from smallest to largest:

0, 1, 3, 3, 4, 5, 6, 7, 8

Notice that there are 9 marks, which means we have an odd number of data points.

The formula for finding the median in this case is:

Median = \dfrac{n + 1}{2} th term.

Here, n = 9. Substituting the value of n gives us:

Median = \dfrac{9 + 1}{2} = \dfrac{10}{2} = 5th term.

The 5th term in our ordered list is 4.

Therefore, the median is 4.

Question 3

The weights (in kg) of 10 students of a class are given below :

21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24. Find the median of their weights.

First, let’s sort the weights in ascending order:

20.5, 21, 21, 22, 24, 24, 25.5, 27.5, 28, 28.5

Notice that there are 10 students, so n = 10, which is an even number.

For an even number of observations, the median is calculated using the formula:

Median = \dfrac{\dfrac{n}{2} \text{ th term} + \Big(\dfrac{n}{2} + 1\Big) \text{ th term}}{2}

Substituting n = 10 into the formula, we have:

= \dfrac{\dfrac{10}{2} \text{th term} + \Big(\dfrac{10}{2} + 1\Big)\text{ th term}}{2} \
= \dfrac{5\text{ th term} + 6\text{ th term}}{2} \
= \dfrac{24 + 24}{2} \
= \dfrac{48}{2} \
= 24.

Hence, median = 24.

Question 4

The marks obtained by 19 students of a class are given below :

27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28. Find:

(i) Median

(ii) lower quartile

(iii) Upper quartile

(iv) Inter-quartile range

First, let’s put the marks in ascending order: 22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 31, 32, 32, 33, 35, 35, 36, 36, 37.

(i) With 19 students, the number of observations, n, is odd.

To find the median, use the formula:

Median = \dfrac{n + 1}{2} th term.

This becomes \dfrac{19 + 1}{2} = \dfrac{20}{2} = 10th term.

Thus, the median is 29.

Hence, median = 29.

(ii) Again, since n = 19 is odd, we calculate the lower quartile:

Lower quartile = \dfrac{n + 1}{4} th term = \dfrac{19 + 1}{4} = \dfrac{20}{4} = 5th term.

So, the lower quartile is 26.

Hence, lower quartile = 26.

(iii) For the upper quartile, use:

Upper quartile = \dfrac{3(n + 1)}{4} th term = \dfrac{3(19 + 1)}{4} = \dfrac{3 \times 20}{4} = 15th term.

Therefore, the upper quartile is 35.

Hence, upper quartile = 35.

(iv) The inter-quartile range is found by subtracting the lower quartile from the upper quartile:

Inter quartile range = Upper quartile – Lower quartile = 35 – 26 = 9.

Hence, inter quartile range = 9.

Question 5

The weight of 60 boys are given in the following distribution table :

Find :

(i) Median

(ii) Lower quartile

(iii) Upper quartile

(iv) Inter quartile range.

To find the required measures, we first need to create a cumulative frequency table:

(i) Since there are 60 boys, n = 60, which is an even number.

The median is calculated using:

Median = \dfrac{\dfrac{n}{2} \text{ th term} + \Big(\dfrac{n}{2} + 1\Big) \text{ th term}}{2}

Substituting the values, we have:

Median = \dfrac{\dfrac{60}{2} \text{ th term} + \Big(\dfrac{60}{2} + 1\Big)\text{ th term}}{2} = \dfrac{\text{30th term + 31st term}}{2}

From the table, the weight for the 30th and 31st boys is 39 kg (as the weights from the 25th to the 42nd term are 39 kg).

∴ Median = \dfrac{39 + 39}{2} = \dfrac{78}{2} = 39 kg.

(ii) For the lower quartile, n = 60 is even.

The formula for the lower quartile is:

Lower quartile = \Big(\dfrac{n}{4}\Big) th term

= \Big(\dfrac{60}{4}\Big) = 15th term.

From the table, the weight for the 15th boy is 38 kg (as the weights from the 11th to the 24th term are 38 kg).

Thus, the lower quartile is 38.

(iii) To find the upper quartile, again n = 60 is even.

The formula for the upper quartile is:

Upper quartile = \Big(\dfrac{3n}{4}\Big) th term

= \Big(\dfrac{3 \times 60}{4}\Big) = 45th term.

From the table, the weight for the 45th boy is 40 kg (as the weights from the 43rd to the 54th term are 40 kg).

Thus, the upper quartile is 40.

(iv) The inter-quartile range is calculated as the difference between the upper and lower quartiles:

Inter quartile range = Upper quartile – Lower quartile

= 40 – 38

= 2.

Therefore, the inter-quartile range is 2.

Question 6

From the following cumulative frequency table draw ogive and then use it to find :

(i) Median

(ii) Lower quartile

(iii) Upper quartile

Let’s organize the data into a cumulative frequency distribution table:

Here, the total number of observations, n, is 80, which is an even number.

For the median, the formula is:

Substituting the values, we have:

For the lower quartile, we use:

Calculating gives:

For the upper quartile, the formula is:

This results in:

Steps to construct the ogive:

1. Use a scale of 1 cm = 10 units on both the x-axis and y-axis.
2. Start by plotting the origin point (0, 0) as the ogive begins at the lower limit of the first class.
3. Plot the points: (10, 5), (20, 24), (30, 37), (40, 40), (50, 42), (60, 48), (70, 70), (80, 77), (90, 79), and (100, 80).
4. Connect these points with a smooth curve.
5. To find the median, draw a horizontal line from the y-coordinate 40 to intersect the curve at point T. From T, draw a vertical line down to the x-axis at point M.
6. For the lower quartile, draw a horizontal line from the y-coordinate 20 to intersect the curve at point O. From O, draw a vertical line down to the x-axis at point P.
7. For the upper quartile, draw a horizontal line from the y-coordinate 60 to intersect the curve at point R. From R, draw a vertical line down to the x-axis at point S.

(i) Observing the graph, the x-coordinate at M is 40.

Hence, median = 40.

(ii) Observing the graph, the x-coordinate at P is 18.

Hence, lower quartile = 18.

(iii) Observing the graph, the x-coordinate at S is 66.

Hence, upper quartile = 66.

Question 7

In a school, 100 pupils have heights as tabulated below :

Find the median height by drawing an ogive.

To handle the given data, we first need to adjust the class intervals to make them continuous. Calculate the adjustment factor using:

Adjustment factor = \frac{\text{Lower limit of one class} - \text{Upper limit of previous class}}{2}

Applying this to our data:

With this, subtract 0.5 from each lower limit and add 0.5 to each upper limit:

Given that n = 100, which is even, we find the median using:

To construct the ogive:

1. Start the x-axis from 120.5, indicating a break or kink at the origin.
2. Use a scale of 2 cm = 10 units on the x-axis.
3. Use a scale of 1 cm = 10 units on the y-axis.
4. Begin plotting at (120.5, 0), as the ogive starts at the x-axis with the lower limit of the first class.
5. Plot the points: (130.5, 12), (140.5, 28), (150.5, 58), (160.5, 78), (170.5, 92), and (180.5, 100).
6. Connect these points with a smooth curve.
7. From the y-axis at 50 (point A), draw a line parallel to the x-axis to meet the curve at point B. From B, draw a line down to the x-axis at point C.

From the graph, point C corresponds to 148 cm.

Thus, the median height is 148 cm.

Question 8

Find the mode of following data, using a histogram :

To determine the mode using a histogram, follow these steps:

1. Begin by constructing a histogram for the provided data.
2. Identify the tallest rectangle, which corresponds to the class with the highest frequency, known as the modal class.
3. From the upper corners of the rectangles adjacent to the modal class, draw two diagonal lines: one from corner C and another from corner D.
4. These diagonals will intersect at a point, labeled as K.
5. From point K, draw a line perpendicular to the horizontal axis, reaching point L.
6. The horizontal coordinate of point L indicates the mode of the data.

∴ The mode is 24.

Hence, mode = 24.

Question 9

The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure.

To determine the mode from the given data, follow these steps:

1. Construct a histogram using the provided frequency distribution.
2. Identify the tallest rectangle in the histogram, which corresponds to the class with the highest frequency, known as the modal class.
3. From the top corners of the rectangles adjacent to the modal class, draw two diagonal lines: one from the upper corner of the rectangle to the left (point M) and the other from the upper corner of the rectangle to the right (point L).
4. The diagonals MJ and LO will intersect at a point Z. From Z, draw a perpendicular line ZP down to the horizontal axis.
5. The point P on the horizontal axis gives the mode of the distribution.

∴ Mode = 34.

Hence, mode = 34.

Question 10

A boy scored the following marks in various class tests during a term, each test being marked out of 20.

15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16.

(i) What are his modal marks ?

(ii) What are his median marks ?

(iii) What are his total marks ?

(iv) What are his mean marks ?

(i) Looking at the given scores, notice that the number 16 appears most frequently.

Hence, mode = 16.

(ii) To find the median, first arrange the scores in increasing order:

7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19.

Here, the total number of scores, n, is 11, which is an odd number.

Using the median formula:

Median = \dfrac{n + 1}{2} th term

= \dfrac{11 + 1}{2} = \dfrac{12}{2}

= 6th term

= 15.

Hence, median = 15.

(iii) To calculate the total marks, sum up all the scores:

Total marks = 7 + 10 + 12 + 12 + 14 + 15 + 16 + 16 + 16 + 17 + 19

= 154.

Hence, total marks = 154.

(iv) The mean is found by dividing the total marks by the number of tests:

Mean = \dfrac{\text{Total marks}}{\text{No. of tests}}

= \dfrac{154}{11}

= 14.

Hence, mean = 14.

Question 11

At a shooting competition the scores of a competitor were as given below :

(i) What was his modal score ?

(ii) What was his median score ?

(iii) What was his total score ?

(iv) What was his mean score ?

(i) Observing the table, the score of 4 occurs most frequently, with a frequency of 7.

Hence, modal score = 4.

(ii) Let’s arrange the scores in a cumulative frequency table:

Here, the total number of shots, n = 25, which is an odd number.

⇒ Median = \dfrac{n + 1}{2} th term = \dfrac{25 + 1}{2} = \dfrac{26}{2} = 13th term.

From the table, the score for the 10th to 13th terms is 3.

Hence, median = 3.

(iii) By summing up the fx column from the cumulative frequency table, we find:

Total score = 80.

Hence, total score = 80.

(iv) Using the formula for mean:

Mean = \dfrac{Σfx}{Σf} = \dfrac{80}{25} = 3.2

Hence, mean = 3.2

Test Yourself

Question 1(a)

The mean of numbers in A.P. 2, 4, 6, 8, ……, 40 is :

• (a) \dfrac{40 + 2}{2}
• (b) \dfrac{20}{2}(2 + 40)
• (c) \dfrac{10}{2} (2 + 40)
• (d) 840

Consider the arithmetic progression (A.P.) given by the sequence 2, 4, 6, 8, …, 40.

Here, the first term a is 2, and the common difference d is calculated as 4 - 2 = 2. The last term l is 40.

To find the number of terms n in this A.P., we use the formula for the n^{th} term:

Substituting the values, we get:

Now, let’s calculate the sum of the first n terms of the A.P. using the formula:

Substituting the known values:

The mean of the A.P. is given by:

To verify, compute:

Thus, the mean is 21. Hence, option 1 is the correct option.

Question 1(b)

The median of 10, 12, 9, 8, 12, 13, 8, 15 and 12 is :

• (a) 12
• (b) \dfrac{8 + 12}{2}
• (c) \dfrac{12 + 13}{2}
• (d) 13

Let’s first arrange the numbers in increasing order: 8, 8, 9, 10, 12, 12, 12, 13, 15.

Now, count the total number of terms, which is 9.

To find the median, use the formula for the position: \dfrac{n + 1}{2}. Here, n = 9, so:

The 5th term in our ordered list is 12.

Hence, Option 1 is the correct option.

Question 1(c)

The numbers 10, 12, 14, 16, 17 and x are in ascending order. If the mean and median of these observations are same, the value of x is :

• (a) 16
• (b) 14
• (c) 54
• (d) 21

We have the numbers 10, 12, 14, 16, 17, and x arranged in ascending order.

The formula for the mean is:

Mean = \dfrac{\text{Sum of observations}}{\text{No. of observations}}

Substituting the numbers, we find:

Since there are 6 numbers, the median is calculated as the average of the 3rd and 4th terms:

Median = \dfrac{\Big(\dfrac{n}{2}\Big) \text{ th term} + \Big(\dfrac{n}{2} + 1\Big) \text{ th term}}{2}

Plugging in the values:

Since it is given that the mean equals the median, we equate the two:

Hence, Option 4 is the correct option.

Question 1(d)

The median of first six prime numbers is :

• (a) 5
• (b) 7
• (c) 6
• (d) 7.5

Consider the first six prime numbers: 2, 3, 5, 7, 11, and 13.

Since there are 6 numbers, which is an even count, the median is calculated by taking the average of the middle two numbers.

The formula for the median when the number of terms is even is:

Plugging in the values, we have:

Hence, Option 3 is the correct option.

Question 1(e)

The inter quartile range for the given ogive is :

1. 42
2. 32
3. 44
4. 54

The total number of terms, denoted as N, is 80.

The lower quartile, Q_1, is calculated as the \dfrac{N}{4}-th term:

Similarly, the upper quartile, Q_3, is determined as the \dfrac{3N}{4}-th term:

To find Q_3, start from point A on the graph, which corresponds to 60. Draw a line parallel to the x-axis until it intersects the graph at point B. Then, from point B, draw a line parallel to the y-axis to meet the graph at point C.

From the graph, observe that point C corresponds to 52.

∴ The upper quartile, Q_3, is 52.

The interquartile range is given by the formula:

Substituting the values, we get:

Hence, Option 1 is the correct option.

Question 1(f)

The mean age of nine boys is 28 years and if one new boy joins them the mean age increases by one.

Assertion(A): The age of new boy is (29 x 10 – 28 x 9) years.

Reason(R): The age of new boy is (29 – 28) x 10 years.

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

The average age of the 9 boys is given as 28 years.

When another boy joins, the average age increases by 1 year, resulting in a new mean of 29 years.

Recall the formula for mean:

Mean = \dfrac{\text{Sum of all observations}}{\text{Number of all observations}}

For the original group of 9 boys:

When the new boy joins, the average for 10 boys becomes 29 years:

The age of the new boy is calculated by finding the difference between the total ages of the 10 boys and the 9 boys:

Thus, the assertion is correct, but the reason given is not. ∴ A is true, R is false.

Hence, option 1 is the correct option.

Question 1(g)

Data = 37, 41, 56, 62, 70, 74, 81, 89, 95 and 90.

Assertion(A): Median = 72.

Reason(R): If number of data(n) is odd, the median = \Big(\dfrac{n + 1}{2}\Big)^{th} term.

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

Let’s start by organizing the data in increasing order: 37, 41, 56, 62, 70, 74, 81, 89, 90, 95. We have 10 data points here.

When the count of numbers, n, is odd, the median is the ( \Big(\dfrac{n+1}{2}\Big)^{th} ) term. However, when n is even, the median is calculated as the average of the ( \Big(\dfrac{n}{2}\Big)^{th} ) and ( \Big(\dfrac{n}{2} + 1\Big)^{th} ) terms.

In this case, n = 10, which is even.

\begin{align}
\text{Median} &= \Bigg(\dfrac{\Big(\dfrac{10}{2}\Big)^{th} + \Big(\dfrac{10}{2} + 1\Big)^{th}}{2}\Bigg) \
&= \Bigg(\dfrac{5^{th} + (5 + 1)^{th}}{2}\Bigg) \
&= \Bigg(\dfrac{5^{th} + 6^{th}}{2}\Bigg) \
&= \Bigg(\dfrac{70 + 74}{2}\Bigg) \
&= \Bigg(\dfrac{144}{2}\Bigg) \
&= 72.
\end{align}

∴ The assertion that the median is 72 is correct. The reason, however, is incorrect because it applies to an odd number of terms, not even. Hence, both the assertion and reason are true, but the reason provided does not justify the assertion.

Hence, option 4 is the correct option.

Question 1(h)

Assertion(A): a = 15 + 25 = 40

b = 50 – a

Reason(R): a + 15 = 25

and b = 50 – 10

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

Cumulative frequency is the running total of frequencies up to a certain class interval. For the interval 10–20, this cumulative frequency must include all frequencies from the previous intervals. Therefore, the cumulative frequency for the 10–20 interval is calculated by adding the frequencies of the 0–10 and 10–20 intervals together.

So, we have:

a = 15 + 25 = 40

For the 20–30 interval, the cumulative frequency given is 50, which includes all previous frequencies. Thus, the frequency for the 20–30 interval is determined by subtracting the cumulative frequency of the previous intervals from 50:

b = 50 – a = 50 – 40 = 10.

∴ A is true, R is false.

Hence, option 1 is the correct option.

Question 1(i)

Data : 9, 11, 15, 19, 17, 13 and 7

Statement (1): For the given data, lower quantile is 11.

Statement (2): For data with n terms, the lower quantile is \Big(\dfrac{n + 1}{4}\Big)^{th} term, if n is odd.

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Consider the dataset: 9, 11, 15, 19, 17, 13, and 7.

First, arrange these numbers in ascending order: 7, 9, 11, 13, 15, 17, 19.

To find the lower quartile for a dataset with an odd number of terms, use the formula for the \Big(\dfrac{n + 1}{4}\Big)^{th} term.

Here, the number of terms, n = 7.

Calculate the lower quartile:

\text{The lower quartile} = \Big(\dfrac{7 + 1}{4}\Big)^{th}
= \Big(\dfrac{8}{4}\Big)^{th}
= 2^{\text{nd term}}
= 9.

Thus, the lower quartile is 9, not 11. This means Statement 1 is incorrect.

Statement 2 correctly describes the method to find the lower quartile when n is odd.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(j)

The mean of given data is 26.

Statement (1): x = 26.

Statement (2): 26 = \dfrac{10 \times 20 + 30 \times x + 50 \times 10}{30 + x}.

• (a) Both the statement are true.
• (b) Both the statement are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Let’s analyze the data:

The mean is calculated using the formula: Mean = \dfrac{\sum f \cdot x}{\sum f}.

Given that the mean is 26, we substitute the values into the formula:

This confirms that statement 2 is correct.

Now, solving the equation:

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(k)

For a given set of data mean = 14 and median = 15.

Statement (1): Mode = 17.

Statement (2): Mode = 3 Median – 2 Mean s

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

We are given that the mean is 14 and the median is 15.

The formula to find the mode in terms of mean and median is:

Mode = 3 \times \text{Median} – 2 \times \text{Mean}

Applying the given values:

⇒ Mode = 3 \times 15 – 2 \times 14

⇒ Mode = 45 – 28

⇒ Mode = 17.

This confirms that Statement 1 is correct because the mode is indeed 17.

Since we used the formula Mode = 3 Median – 2 Mean to find the mode, Statement 2 is also correct.

∴ Both statements are true.

Hence, option 1 is the correct option.

Question 2

The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m – 1 and median q. Find p and q.

We start with the information that the mean of the numbers 1, 7, 5, 3, 4, and 4 is denoted by m.

First, calculate the sum of these numbers: 1 + 7 + 5 + 3 + 4 + 4 = 24.

The mean, m, is then calculated as:
\text{Mean} (m) = \dfrac{\text{Sum of observations}}{\text{No. of observations}} = \dfrac{24}{6} = 4.

Next, consider the numbers 3, 2, 4, 2, 3, 3, and p, which have a mean of m – 1, or 3.

The sum of these numbers is given by 3 + 2 + 4 + 2 + 3 + 3 + p = 17 + p.

Using the formula for the mean, we have:
3 = \dfrac{17 + p}{7}
Solving for p, we multiply both sides by 7:
21 = 17 + p
Thus, p = 4.

Now, arrange the numbers 2, 2, 3, 3, 3, 4, and 4 in ascending order.

With n = 7, which is an odd number, the median is found using the formula:
\text{Median} = \dfrac{n + 1}{2}\text{th term}
Substitute the value of n:
\dfrac{7 + 1}{2} = \dfrac{8}{2} = 4\text{th term}
The 4th term in the ordered list is 3.

∴ q = 3.

Hence, p = 4 and q = 3.

Question 3

In a malaria epidemic, the number of cases diagnosed were as follows :

On what days do the mode, the upper and the lower quartiles occur ?

Let’s create a cumulative frequency distribution table to better understand the data:

Notice that the total number of cases, n, is 206, which is an even number.

To find the lower quartile, we calculate the \dfrac{n}{4} th term:

Lower quartile = \dfrac{206}{4} = 51.5 th term

From the table, we observe that the 51.5th term falls on the date range from the 38th to the 64th term, which corresponds to 4th July.

Next, for the upper quartile, we determine the \dfrac{3n}{4} th term:

Upper quartile = \dfrac{3 \times 206}{4} = 154.5 th term

From the table, the 154.5th term is within the range of the 141st to the 171st term, which is 7th July.

Finally, by examining the table, we see that the highest number of cases, 46, occurred on 5th July.

Hence, mode = 5th July, upper quartile = 7th July, and lower quartile = 4th July.

Question 4

The marks obtained by 120 students in a Mathematics test are given below :

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate :

(i) the median

(ii) the number of students who obtained more than 75% marks in a test ?

(iii) the number of students who did not pass in the test if the pass percentage was 40?

(iv) the lower quartile.

Cumulative Frequency Distribution Table:

(i) Constructing the Ogive:

1. Use a scale of 1 cm = 10 marks on the x-axis.
2. Use a scale of 1 cm = 20 students on the y-axis.
3. Start the ogive from the origin (0, 0), as it represents the lower limit of the first class.
4. Plot the cumulative frequency points: (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), and (100, 120).
5. Connect these points with a smooth curve.
6. Draw a horizontal line from the point where the cumulative frequency is 60 to intersect the ogive at point B. From point B, draw a vertical line down to the x-axis, reaching point C.

From the graph, point C is at 43.

Thus, the median is 43.

(ii) Students Scoring More Than 75%:

Total marks = 100.

Calculate 75% of 100 marks: \dfrac{75}{100} \times 100 = 75.

Draw a vertical line from 75 on the x-axis to meet the curve at point E. From point E, draw a horizontal line to the y-axis, reaching point F.

From the graph, point F is at 110.

This indicates 110 students scored 75% or less.

Thus, the number of students scoring more than 75% is 120 - 110 = 10.

Therefore, 10 students scored more than 75%.

(iii) Students Failing the Test:

Total marks = 100.

Calculate 40% of 100 marks: \dfrac{40}{100} \times 100 = 40.

Draw a vertical line from 40 on the x-axis to meet the curve at point H. From point H, draw a horizontal line to the y-axis, reaching point I.

From the graph, point I is at 52.

Thus, 52 students did not pass.

(iv) Finding the Lower Quartile:

Given n = 120, which is even.

The formula for the lower quartile is \dfrac{n}{4} = \dfrac{120}{4} = 30th term.

Draw a horizontal line from the 30th student mark on the y-axis to meet the curve at point K. From point K, draw a vertical line down to the x-axis, reaching point L.

From the graph, point L is at 30.

Therefore, the lower quartile is 30.

Question 5

Using a graph paper, draw an ogive for the following distribution which shows a record of the weight in kilograms of 200 students.

Use your ogive to estimate the following :

(i) The percentage of students weighing 55 kg or more.

(ii) The weight above which the heaviest 30% of the students fall,

(iii) The number of students who are (a) under-weight and (b) over weight, if 55.70 kg is considered as standard weight ?

(i) Let’s create a cumulative frequency table:

Steps to draw the ogive:

1. Begin the x-axis from 40 with a break, as the scale starts at this point.
2. Use 1 cm on the x-axis to represent 5 kg.
3. Use 1 cm on the y-axis to represent 20 units.
4. Mark the starting point (40, 0) on the graph, as the ogive starts here.
5. Plot the points: (45, 5), (50, 22), (55, 44), (60, 89), (65, 140), (70, 171), (75, 191), and (80, 200).
6. Connect these points smoothly with a freehand curve.
7. From the weight 55 kg on the x-axis, draw a vertical line to meet the curve at point Q. Then, draw a horizontal line from Q to the y-axis at point K.

From the graph, K = 44.

This means 44 students weigh 55 kg or less.

∴ Students weighing more than 55 kg = 200 – 44 = 156.

Percentage of students weighing more than 55 kg = \dfrac{156}{200} \times 100 = 78%.

Hence, percentage of students weighing more than 55 kg = 78%.

(ii) Calculate 30% of students:

30% of students = \dfrac{30}{100} \times 200 = 60.

Total students = 200

Students not in the heaviest 30% = 200 – 60 = 140.

From 140 on the y-axis, draw a horizontal line to meet the curve at point R, then a vertical line from R to the x-axis at point P.

From the graph, P = 65

Hence, above 65 kg the heaviest 30% of the students fall.

(iii) To find under-weight and over-weight students:

Draw a vertical line from 55.70 kg on the x-axis to the curve at point M. Then, draw a horizontal line from M to the y-axis at point N.

(a) From the graph, N = 46.

∴ 46 students weigh less than 55.70 kg

Hence, 46 students are underweight.

(b) Since 46 students weigh less than 55.70 kg,

∴ 154 (200 – 46) students weigh more than 55.70 kg

Hence, 154 students are overweight.

Question 6

The distribution given below, shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

To analyze the data, let’s first construct a cumulative frequency distribution table:

To find the mean, apply the formula:

Mean = \dfrac{Σfx}{Σf} = \dfrac{171}{25} = 6.84

Given that the total number of students, n, is 25, which is odd, the median is determined by finding the \dfrac{n + 1}{2}th term.

Median = \dfrac{25 + 1}{2} = \dfrac{26}{2} = 13th term.

Referring to the table, the 13th student falls in the group with marks obtained as 7.

∴ Median = 7.

For the mode, observe that the highest frequency is 9, corresponding to marks of 6.

∴ Mode = 6.

Hence, mean = 6.84, median = 7 and mode = 6.

Question 7

The monthly income of a group of 320 employees in a company is given below :

Draw an ogive of the given distribution on a graph sheet taking 2 cm = ₹ 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine :

(i) the median wage.

(ii) the number of employees whose income is below ₹ 8500.

(iii) if the salary of a senior employee is above ₹ 11500, find the number of senior employees in the company.

(iv) the upper quartile.

(i) Let’s organize the data into a cumulative frequency table:

For 320 employees, n = 320, which is an even number.

The median is found using:

Median = \dfrac{n}{2} th term

= \dfrac{320}{2} = 160th term.

Here’s how to plot the ogive:

1. The x-axis begins at 6, so a break is shown near the origin to indicate the graph starts at this point.
2. Use a scale of 2 cm = ₹ 1000 on the x-axis.
3. Use a scale of 1 cm = 40 employees on the y-axis.
4. Plot the starting point (6, 0) for the ogive.
5. Plot the cumulative points: (7, 20), (8, 65), (9, 130), (10, 225), (11, 285), (12, 315), and (13, 320).
6. Connect these points smoothly with a freehand curve.
7. From the 160th employee mark on the y-axis, draw a horizontal line to intersect the ogive at point N. Then, draw a vertical line from N to the x-axis at point O.

From the graph, point O corresponds to 9.2 (thousands).

Thus, the median wage is ₹ 9200.

(ii) For incomes below ₹ 8500:

Draw a vertical line from ₹ 8.5 (thousands) on the x-axis to meet the ogive at point Q. From Q, draw a horizontal line to the y-axis at point R.

According to the graph, point R is at 95.

Therefore, 95 employees earn less than ₹ 8500.

(iii) For salaries above ₹ 11500:

Draw a vertical line from ₹ 11.5 (thousands) on the x-axis to meet the ogive at point T. From T, draw a horizontal line to the y-axis at point U.

The graph indicates that point U is at 305.

∴ 305 employees earn less than ₹ 11500.

∴ 15 employees (320 – 305) earn more than ₹ 11500.

Hence, there are 15 senior employees.

(iv) To find the upper quartile:

For n = 320, which is even, use:

Upper quartile = \dfrac{3n}{4} = \dfrac{3 \times 320}{4} = 240th term.

Draw a horizontal line from the 240th employee mark on the y-axis to intersect the ogive at point W. Then, draw a vertical line from W to the x-axis at point Z.

From the graph, point Z corresponds to 10.3 (thousands).

Therefore, the upper quartile is ₹ 10300.

Question 8

The mean of numbers 45, 52, 60, x, 69, 70, 26, 81 and 94 is 68. Find the value of x. Hence, estimate the median for the resulting data.

To determine the value of x, we first calculate the sum of all the numbers: 45, 52, 60, x, 69, 70, 26, 81, and 94. This gives us a total of 497 + x.

The total number of observations is 9.

Using the formula for the mean:

Mean = \dfrac{\text{Sum of observations}}{\text{No. of observations}}

we substitute the given mean value:

68 = \dfrac{497 + x}{9}

Multiplying both sides by 9, we have:

612 = 497 + x

Solving for x gives:

x = 612 – 497 = 115.

Now, to find the median, note that the number of observations n = 9 is odd. The median is the middle value, which is the \dfrac{n + 1}{2}th term.

Median position = \dfrac{9 + 1}{2} = \dfrac{10}{2} = 5th term.

The 5th term in the ordered list is 69.

Hence, mean = 115 and median = 69.

Question 9

The marks of 10 students of a class in an examination arranged in ascending order is as follows :

13, 35, 43, 46, x, x + 4, 55, 61, 71, 80.

If the median marks is 48, find the value of x. Hence, find the mode of the given data.

We have a total of 10 students’ marks, so n = 10, which is an even number.

To find the median, we use the formula:

Substituting the given values:

This simplifies to:

In the data set, the 5th term is x and the 6th term is x + 4. Therefore:

Simplifying further:

We know that the median is given as 48.

⇒ x + 2 = 48

⇒ x = 46.

Now, substituting x = 46 back into the data set, we have: 13, 35, 42, 46, 46, 50, 55, 61, 71, 80.

Notice that 46 appears most frequently in this list.

∴ The mode is 46.

Hence, x = 46 and mode = 46.

Question 10

The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to :

(i) Frame a frequency distribution table.

(ii) To calculate mean.

(iii) To determine the modal class.

(i) Let’s organize the data into a frequency distribution table:

(ii) To find the mean, we first compute the class mean and then the product of frequency and class mean (fx) for each class:

Using the formula for mean:

Mean = \dfrac{Σfx}{Σf}

= \dfrac{695}{25} = 27.8

Thus, the mean is 27.8.

(iii) Observing the frequency distribution table, the class interval with the highest frequency is 20 – 30.

Therefore, the modal class is 20 – 30.