Marks of 30 students are listed in the ungrouped distribution as given below.

18, 32, 54, 67, 23, 1, 23, 21, 10, 19, 81, 17, 24, 12, 45, 65, 67, 80, 23, 12, 11, 12, 21, 3, 10, 45, 47, 34, 85, 43.

If this distribution is converted into grouped frequency distribution, what is the length of class interval if number of class interval is 7 ?

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Solution

The correct option is D 12
Range of the given dataset = Maximum value - Minimum value
= 85 - 1 = 84

Number of class interval given = 7.

$Length of class interval = \frac{R a n g e}{Number of class interval}$
= $\frac{84}{7}$ = 12

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Similar questions

Marks of 30 students are listed in the ungrouped distribution as given below. 13, 32, 45, 67, 23, 18, 23, 21, 10, 19, 18, 17, 24, 12, 45, 56, 76, 80, 23, 12, 11, 12, 19, 15, 19, 54, 74, 34, 58, 34.

If this distribution is converted into a grouped frequency distribution, what is the length of the class interval if the number of the class interval is 5?

Q. The marks obtained by

40

students of a class in an examination are given:

8, 47, 22, 31, 17, 13, 38, 26, 3, 34, 29, 11, 22, 7, 15, 24, 38, 31, 21, 35, 42, 24,

45, 23, 21, 27, 29, 49, 25, 48, 21, 15, 18, 27, 19, 45, 14, 34, 37, 34

Prepare a frequency distribution table with equal class intervals, starting from

0 - 10

[where

10

is not included].

Q. Given below are the marks obtained by

30

students in an examination:

$08$	$17$	$33$	$41$	$47$	$23$	$20$	$34$
$09$	$18$	$42$	$14$	$30$	$19$	$29$	$11$
$36$	$48$	$40$	$24$	$22$	$02$	$16$	$21$
$15$	$32$	$47$	$44$	$33$	$01$

Taking class intervals

1 - 10, 11 - 20 . . . . . . . . . . . . .41 - 50;

make a frequency table for the above distribution.

Q. Marks of the 100 students are distributed in the grouped frequency distribution table.

Class marks (Marks)	150	160	170	180	190	200
No. of Students	23	14	16	11	21	15

What is the class interval for the group having frequency 11 ?

Q. The marks of 24 candidates in the subject mathematics are given below:

45	48	15	23	30	35	40	11
29	0	3	12	48	49	18	30
15	30	11	42	23	2	3	44

The maximum marks are 50. Make a frequency distribution taking class intervals 0 – 10, 10-20…

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The correct option is D 12Range of the given dataset = Maximum value - Minimum value = 85 - 1 = 84 Number of class interval given = 7. Length of class interval=RangeNumber of class interval = 847 = 12

The correct option is D 12
Range of the given dataset = Maximum value - Minimum value
= 85 - 1 = 84

Number of class interval given = 7.

$Length of class interval = \frac{R a n g e}{Number of class interval}$
= $\frac{84}{7}$ = 12