Question - 11 : -
An incomplete distribution is given below :

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.

Answer - 11 : -

Let p₁, and p₂ be the missing frequencies

Median = 46 and N = 230

∴ 150 +p₁ + p₂ = 230

⇒ p₁+p₂ =230 – 150 = 80
∴ p₂ = 80-p₁ ….(i)
∵ Median = 46 which lies inthe class 40-50
∴ I = 40, f= 65, F = 42 +p₁,h = 10

⇒ 39 = 73 – p₁
⇒ p₁ = 73-39 = 34
∴ p₂ – 80 –p₁ = 80 – 34 = 46
Hence missing frequencies are 34, and 46
Mean Let assumed mean (A) = 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:

Answer - 12 : - Mean = 28.5 , N = 60

17 = 25 -f₁
⇒ f₁=25 -17 = 8
and f₂ = 15-f₁ = 15-8 = 7
Hence missing frequencies are 8 and 7

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:

Answer - 13 : - Median = 525, N = 100

⇒525 – 500 = (14 -f₁) x 5

⇒ 25 = 70- 5f₁
⇒ 5f₁ =70 – 25 = 45
⇒ f₁ = 45/5 =9
and f₂ = 24 – f₁ = 24 – 9 = 15
Hence f₁ = 9, f₂ = 15

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

Answer - 14 : - Mean = 32.5 and N= 40

⇒ 2.5 x 12 = 60 – 10f₁
⇒ 30 =60 – 10f₁
⇒ 10f₁ =60-30 = 30
⇒ f₁ = 30/10 =3
∴ f₂ =9 – f₁ = 9-3 = 6
Hence f₁ = 3, f₂= 6

Question - 15 : - Compute the median for each of the following data:

Answer - 15 : -

Here N= 100=

∴ N/2 = 100/2 = 50 which lies in the class 70-90 (∵ 50 < 65 and > 43)

∴ l = 70, F =43 , f = 22 ,h = 20

(ii) Greater than

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

∴ l = 110, F = 60 , f=45, h= 10

Question - 16 : -
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and
the following data was obtained.

Find the median height.

Answer - 16 : -

Here N/2 = 51/2 = 25.5 or 26 which lies in the class 145-150

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.

Answer - 17 : -

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

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

= 35 + 0.76 = 35.76

Question - 18 : - The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
Find the mean length of leaf.

Answer - 18 : -

N =40, N/2 = 40/2 =20 which lies in the class 144.5-153.5 as 17 < 20 < 29
∴ l=144.5, F= 17, f= 12, h = 9

= 144.5 + 2.25 = 146.75

Question - 19 : -
An incomplete distribution is given as follows :

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.

Answer - 19 : - Median = 25 and ∑f= N = 170
Let p₁ and p₂ be two missing frequencies

∴ 110 + p₁ +p₂ = 170

⇒ p₁+ p₂ = 170 – 110 = 60

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

∴ Median = 35 which lies in the class 30-40

Here l = 30, f= 40, F = 30 + p₁ and h = 10

20 = 55 – p₁

⇒ p₁ = 55 – 20 = 35

But p₁+ p₂ = 60

∴ p₂ = 60 -p₁ = 60 – 35 = 25

Hence missing frequencies 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.

Answer - 20 : -

It is given that n = 20
So, 10 + x + y – 20, i.e., x+y= 10 …(i)
It is also given that median = 14.4
Which lies in the class interval 12-18
So, l = 12, f= 5, cf = 4 + x, h = 6
Using the formula,

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