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Assumed Mean Method of Finding Mean

Find the mean...

Question

Find the mean of the following data

Marks

1 - 10

11 - 20

21 30

31 - 40

41 -50

51 - 60

61 - 70

No. of student

6

5

8

15

7

6

3

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Solution

$x_{i} = 5, 15, 25, 35, 45, 55, 65$
$\sum f_{i} = 50$
$\sum f_{i} x_{i} = 30 + 75 + 200 + 525 + 315 + 330 + 195 = 1670$
$¯ x = \frac{\sum f_{i} x_{i}}{\sum x_{i}} = \frac{1670}{50} = 33.4$

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Q. Calculate the mean of the distribution given below using the short cut method.

Marks	$11 - 20$	$21 - 30$	$31 - 40$	$41 - 50$	$51 - 60$	$61 - 70$	$71 - 80$
No. of students	$2$	$6$	$10$	$12$	$9$	$7$	$4$

Marks	No. of students
$1 - 20$	$8$
$21 - 40$	$12$
$41 - 50$	$15$
$51 - 60$	$20$
$61 - 70$	$13$
$71 - 80$	$17$
$81 - 90$	$5$

Number of students obtained marks between

41

80

40

bits of wire, correct to the nearest centimetre are given below. Calculate the variance.

Length cm	$1 - 10$	$11 - 20$	$21 - 30$	$31 - 40$	$41 - 50$	$51 - 60$	$61 - 70$
No. of bits	$2$	$3$	$8$	$12$	$9$	$5$	$1$

Marks	No. of students
$1 - 20$	$8$
$21 - 40$	$12$
$41 - 50$	$15$
$51 - 60$	$20$
$61 - 70$	$13$
$71 - 80$	$17$
$81 - 90$	$5$

Write the probability of a student obtain less than

51

Marks	No. of students
$1 - 20$	$8$
$21 - 40$	$12$
$41 - 50$	$15$
$51 - 60$	$20$
$61 - 70$	$13$
$71 - 80$	$17$
$81 - 90$	$5$

find probability that

A student obtained more than

50

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