NCERT Solutions
Class 10 Maths
Statistics
Ex.13.3 Q.6
100 surnames were randomly picked up from a local telephone directory and the
frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Determine the median number of letters in the surnames.
Find the mean number of letters in the surnames?
Also, find the modal size of the surnames.
The cumulative frequencies with their respective class intervals are as follows:
It can be observed that the cumulative frequency just greater than 
(i.e., 100 ÷ 2 = 50) is 76,
belonging to class interval 7 − 10.
Median class = 7 − 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
Median = l + {( – cf) ÷ f} × h
= 7 + {(50 - 36) ÷ 40} × 3
= 7 + (14 ÷ 40) × 3
= 7 + 
= 7 + 1.05
= 8.05
To find the class mark (xi) for each interval, the following relation is used.
Class mark xi = (Upper limit + Lower limit) ÷ 2
Taking 11.5 as assured mean (a), di, ui, and fiui can be calculated according to step deviation method as follows:
From the table, it can be observed that
Σfi = 100
Σfiui = -106
Now, Mean = a + (Σfiui ÷ Σfi ) h
= 11.5 + (-106 ÷ 100) × 3
= 11.5 - 
= 11.5 - 3.18
= 8.32
The data in the given table can be written as:
From the table, it can be observed that the maximum class frequency is 40 belonging to class
interval 7 − 10. Therefore, modal class = 7 − 10
Lower limit (l) of modal class = 7
Class size (h) = 3
Frequency (f1) of modal class = 40
Frequency (f0) of class preceding the modal class = 30
Frequency (f2) of class succeeding the modal class = 16
Mode = l + {(f1 – f0) ÷ (2f1 – f0 – f2)} × h
= 7 + {(40 - 30) ÷ (2 × 40 – 30 – 16)} × 3
= 7 + 10 ÷ (80 - 46) × 3
= 7 + 
= 7 + 0.88
= 7.88
Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively
while modal size of surnames is 7.88.