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Stastics — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsStastics5 Marks
Question
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:
Number of letters 1-4 4-7 7-10 10-13 13-16 16-19
Number of surnames 6 30 40 16 4 4
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.
✓

Answer

First, we will convert the graph given into tabular form as shown below:
Class interval Frequency ($f_i$) Mid value ($x_i​​​​​​​$) $f_ix_i$ Cumulative Frequency
1 – 4 6 2.5 15 6
4 – 7 30 5.5 165 36
7 – 10 40 8.5 340 76
10 – 13 16 11.5 184 92
13 – 16 4 14.5 58 96
16 – 19 4 17.5 70 100
  N = $\sum$$f_i$ = 100   $\Sigma f_i x_i$ = 832  
  1. N = 100
    Mean = $\frac { \Sigma f_i x_i } { N } = \frac { 832 } { 100 }$ = 8.32
  2. $ \frac{N}{2} = \frac{{100}}{2}$ = 50
    The cumulative frequency just greater than $\frac {N} {2}$ is 76, then the median class is 7 - 10 such that
    l = 7, h = 10 - 7 = 3, f = 40, F = 36
    Median = l + $\frac { \frac { N } { 2 } - F } { f } \times$ h
    = 7 + $\frac { 50 - 36 } { 40 } \times$ 3
    $= 7 + \frac {42} {40}$ = 7 + 1.05 = 8.05
  3. Mode = 3 Median - 2 Mean
    = 3 $\times$ 8.05 - 2 $\times$ 8.32 = 7.51

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