Question 11
Following is the distribution of marks obtained by 60 students in science test
| Marks | No.of Students |
| More than 0 | 60 |
| More than 10 | 56 |
| More than 20 | 40 |
| More than 30 | 20 |
| More than 40 | 10 |
| More than 50 | 3 |
Solution
Mean = 26.5
Question 12
If mode of data is 23/7 , then find x and y
| Class | 1-3 | 3-5 | 5-7 | 7-9 | 9-11 | Total |
| Frequency | X | 8 | Y | 2 | 1 | 20 |
Solution
X= 7 and y= 2
Question 13
Bowling strike rate for a bowler is defined as the average number of balls bowled per wicket
taken. A bowler has taken 148 wickets so far with a strike rate of 27. In his next match, he
bowls 48 balls and takes 2 wickets.
What is his new strike rate? Show your work.
Solution
Find the total number of balls bowled by the baller so far as 148×27=3996
Finds the total number of balls bowled after the last match as 3996+48=4044.
and the total number of wickets taken after the latest match as 148+2=150.
Find the new strike rate of the bowler as (total number of balls bowled)/(total number
of wickets taken) = 4044/150 = 26.96
Question 14
The frequency distribution of daily rainfall in a town during a certain period is shown below
| Rainfall (in mm) | Number of days |
| 0-20 | 7 |
| 20-40 | X |
| 40-60 | 10 |
| 60-80 | 4 |
Unfortunately, due to manual errors, the information on the 20-40 mm range got deleted from
the data.
If the mean daily rainfall for the period was 35mm, find the number of days when the rainfall
ranged between 20-40 mm. Show your work
Solution
the value of x as 23.
Question 15
If the mean of the following distribution is 6, find the value of‟p‟.
| x | 2 | 4 | 6 | 10 | P+6 |
| y | 3 | 2 | 3 | 1 | 2 |
Solution
. P = 7
Question 16
Find the median of the following data:
| Wages(Rs) | No of workers |
| More than 150 | nill |
| More than 140 | 12 |
| More than 130 | 27 |
| More than 120 | 60 |
| More than 110 | 105 |
| More than 100 | 124 |
| More than 90 | 141 |
| More than 80 | 150 |
Solution
= 116.67
Question 17
If mode of the following frequency distribution is 55, then find the value of x
| Class | Frequency |
| 0-15 | 10 |
| 15-30 | 7 |
| 30-45 | X |
| 45-60 | 15 |
| 60-75 | 10 |
| 75-90 | 12 |
Solution
Mode=55 , So Modal class= 45-60
Mode = l +[(f1-f0)/(2f1 –f0 –f2 )] x h
Or,55 =45 + 15x(15-x)/(2×15-x-10 )
Or, 55 = 45 + (225- 15x)/ 30-x-10
Or, 55(20-x) = 45(20-x) +225-15x
Or,1100 – 55x = 900 -45x +225 -15x
Or, 5x=1125-1100
x=5
SELF ASSESSMENT
Q1. The mean of the following frequency distribution is 25. Find the value of f
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 5 | 18 | 15 | f | 6 |
Q2. The mean of the following data is 18.75. Find the value of P
| Class Mark | Frequency |
| 10 | 5 |
| 5 | 10 |
| P | 7 |
| 25 | 8 |
| 30 | 2 |
Q3. Following table shows sale of shoes in a store during one month
| Size of shoes | Pairs of shoes sold |
| 3 | 4 |
| 4 | 18 |
| 5 | 25 |
| 6 | 12 |
| 7 | 5 |
| 8 | 1 |
