# Class 10 RD Sharma Solution – Chapter 7 Statistics – Exercise 7.4 | Set 2

### You are given that the median value is 46 and the total number of items is 230.

(i) Using the median formula fill up missing frequencies.

(ii) Calculate the AM of the completed distribution.

Solution:

Let us assume p1, and p2 to be the missing frequencies

Median = 46 and N = 230

âˆ´ 150 + p1 + p2 = 230

â‡’ p1+p2 = 230 â€“ 150 = 80

âˆ´ p2 = 80-p1 â€¦..(i)

Since, median = 46 which lies in the class interval belonging to 40-50

âˆ´ I = 40, f= 65, F = 42 +p1, h = 10

â‡’ 39 = 73 â€“ p1

â‡’ p1 = 73 -39 = 34

âˆ´ p2 â€“ 80 â€“ p1 = 80 â€“ 34 = 46

Therefore, the missing frequencies are 34, and 46.

Let the assumed mean (A) be 45.

= 45 + 0.8695

= 45 + 0.87

= 45.87

### Question 12. If the median of the following frequency distribution is 28.5 find the missing frequencies:

Solution:

Mean = 28.5, N = 60

Therefore,

45 + f1 + f2 = 60

=> f1 + f2 = 60 – 45 = 15

=> f2 = 15 – f1

17 = 25 – f1

N/2 = 30

Now, Median = 28.5 and it lies in the class interval of 20-30

Therefore,

l = 20, F = 5 + f1, f= 20 and h = 10

â‡’ f1= 25 -17 = 8

and f2 = 15-f1 = 15-8 = 7

Therefore, the missing frequencies are 8 and 7 respectively.

### Question 13. The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:

Solution:

Median = 525, N = 100

Therefore,

76 + f1 + f2 = 100 => f1 + f2 = 100 – 76 = 26

f = 24 – f1

Because,

Median = 525 which belongs to the interval 500-600

Now, l =500, F = 36 + f1, f =20, h = 100

Therefore,

M

â‡’ 525 â€“ 500 = (14 -f1) x 5

â‡’ 25 = 70- 5f1

â‡’ 5f1 = 70 â€“ 25 = 45

â‡’ f1 = 455 = 9

and f2 = 24 â€“ f1 = 24 â€“ 9 = 15

Hence, we obtain the values for f1 = 9, f2 = 15.

### Question 14. If the median of the following data is 32.5, find the missing frequencies.

Solution:

Mean = 32.5 and N= 40

Now, we know,

Solving, we get,

â‡’ 2.5 x 12 = 60 â€“ 10f1

â‡’ 30 = 60 â€“ 10f1

â‡’ 10f1 = 60-30 = 30

â‡’ f1 = 30/10 =3

âˆ´ f2 = 9 â€“ f1 = 9-3 = 6

Hence, f1 = 3, f2= 6

### Question 15. Compute the median for each of the following data:

(i)

Solution:

(i) Less than

We have, N= 100

âˆ´N/2 = 100/2 = 50 which lies in the class  interval belonging to 70-90 (âˆµ 50 < 65 and > 43)

âˆ´ l = 70, F =43 , f = 22 ,h = 20

(ii) Greater than

We have,

N = 150, N/2 = 150/2 = 75 which lies in the class interval belonging to 110-120 (âˆµ 75 > 105 and 75 > 60)

âˆ´ l = 110, F = 60 , f=45, h= 10

### Find the median height.

Solution:

Here, âˆ‘F/2 = 51/2 = 25.5 or 26 which lies in the class interval belonging to 145-150

Therefore,

l= 145, F= 11, f= 18, h= 5

= 145 + 4.03 = 149.03

### Question 17. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years.

Solution:

Here N = 100, N/2 = 100/2 = 50 which lies in the class interval of 35-40 ( âˆµ 45 > 50> 78)

Therefore,

l = 35, F = 45, f= 33, h = 5

= 35 + 0.76 = 35.76

### Find the mean length of leaf.

Solution:

N = 40, N/2 = 40/2 = 20 which lies in the class interval of 144.5-153.5 as 17 < 20 < 29

Therefore,

l= 144.5, F= 17, f= 12, h = 9

= 144.5 + 2.25 = 146.75

### You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.

Solution:

Median = 25 and âˆ‘f= N = 170

Let us assume x and y to be the two missing frequencies

âˆ´ 110 + x +y = 170

â‡’ x + y = 170 â€“ 110 = 60

Here, we have,

N = 170, N/2 = 170/2 = 85

Therefore, Median = 35 which lies in the class interval belonging to 30-40

Here l = 30, f= 40, F = 30 + x and h = 10

20 = 55 â€“ x

â‡’ x = 55 â€“ 20 = 35

But,

x + y = 60

Solving for y, we get,

âˆ´ y = 60 – x = 60 â€“ 35 = 25

Hence missing frequencies x and y are 35 and 25.

### Question 20. The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.

Solution:

We know, n = 20

Therefore,

10 + x + y â€“ 20,

=> x+y= 10 â€¦(i)

Also,

Median = 14.4 which lies in the class interval belonging to 12-18

So, l = 12, f= 5, cf = 4 + x, h = 6

Solving for x, we get,

x = 6 ….(ii)

Also,

y = 6

### Question 21. The median of the following data is SO. Find the values of p and q, if the sum of all the frequencies is 90.

Solution:

Given, N = 90

And, N/2 = 90/2 = 45 which lies in the class interval 50-60

Now,

Lower limit, l = 50, f= 20, cf= 40 + p, h = 10

Obtaining values, we get,

âˆ´ P = 5

Also, 78 +p + q = 90

â‡’ 78 + 5 + q = 90

â‡’ q = 90-83

âˆ´ q = 7

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