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Home » In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:
In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:
CBSE | Class 10 | Exercise 14.3 | Maths | NCERT | Statistics
In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:
Number of letters1-44-77-1010-1313-1616-19
Number of surnames630401644

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution:

To compute middle:

Class Interval      Frequency           Cumulative Frequency

1-4         6             6

4-7         30           36

7-10       40           76

10-13     16           92

13-16     4             96

16-19     4             100

Given:

n = 100 &n/2 = 50

Middle class = 7-10

Consequently, l = 7, Cf = 36, f = 40 and h = 3

Ncert solutions class 10 chapter 14-6

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> Middle\text{ }=\text{ }7+\left( \left( 50-36 \right)/40 \right)\text{ }\times \text{ }3 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> ~ \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

    \[Middle\text{ }=\text{ }7+42/40\]

    \[Median=8.05\]

Compute the Mode:

Modular class = 7-10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 and h = 3

Ncert solutions class 10 chapter 14-7

    \[Mode\text{ }=\text{ }7+\left( \left( 40-30 \right)/\left( 2\times 40-30-16 \right) \right)\text{ }\times \text{ }3\]

    \[=\text{ }7+\left( 30/34 \right)\]

= 7.88

Consequently mode = 7.88

Compute the Mean:

Class Interval      fi             xi            fixi

1-4         6             2.5         15

4-7         30           5.5         165

7-10       40           8.5         340

10-13     16           11.5       184

13-16     4             14.5       51

16-19     4             17.5       70

Total fi = 100                     Sum fixi = 825

    \[Mean\text{ }=\text{ }x\text{ }=\text{ }\sum fi\text{ }xi/\sum fi\]

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph --> Mean\text{ }=\text{ }825/100\text{ }=\text{ }8.25 \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> ~ \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

Consequently, mean = 8.25

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