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Worked Examples · Example 39
Q.

In a test of 50 students, marks secured by students is given in the following table. Calculate the standard deviation of scores.

Height (inch)10-2020-3030-4030-4040-50
Number of students5815166

(Note: the class-interval row is printed exactly as in the book, with "30-40" repeated; the intended first interval is very likely "0-10" — flagged as a probable misprint.)

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The book's class-interval row prints "30-40" twice; since the class frequencies total the stated 5050 students only if the first class is 00-1010, that correction is used here. With it, the mean is 2727 marks and the standard deviation is ≈11.49\approx11.49 marks.

xˉ=∑fixiN\bar x=\dfrac{\sum f_ix_i}{N}, σ2=∑fixi2N−xˉ2\quad \sigma^2=\dfrac{\sum f_ix_i^2}{N}-\bar x^2 (the shortcut/product-sum form of variance), σ=σ2\quad \sigma=\sqrt{\sigma^2}

  1. Note on the printed data: the class-interval row is given as "30-40" twice with no "0-10" class. Since the frequencies 5,8,15,16,65,8,15,16,6 must sum to the 5050 students stated in the problem (and they do: 5+8+15+16+6=505+8+15+16+6=50), the intended first interval is taken as 00-1010, per the flagged probable misprint.
  2. Working table:
CI0-1010-2020-3030-4040-50Total
mid xix_i515253545
fif_i5815166N=50N=50

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