In order to draw the histogram of the inclusive frequency distribution, it should be converted into exclusive form by applying a correction factor. A correction factor is calculated as

\frac{d}{2}

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\frac{m}{2}

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\frac{n}{2}

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\frac{l}{2}

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Solution

The correct option is A $\frac{d}{2}$
The given distribution is in inclusive form. It should be converted into exclusive form. This can be done by applying a correction factor $\frac{d}{2}$ , where

d = (lower limit of a class) - (upper limit of a class before it)

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While drawing a histogram for the discontinuous grouped data, if in case, the class intervals are given in the inclusive form, then in order to convert them into exclusive form, we must,

Q. Marks scored by 50 students are given below:

40	45	38	24	46	42	45	18	53	64
45	32	52	54	78	65	52	64	66	43
48	55	50	43	48	20	27	65	37	55
51	55	62	66	38	16	60	58	46	35
72	62	54	58	30	36	43	82	46	53

(a) Arrange the marks in ascending order.
(b) Represent the marks in the form of discrete frequency distribution.
(c) Construct an inclusive frequency distribution with first class as 10−19. Also construct class boundaries.
(d) Construct a frequency distribution with exclusive class-intervals, taking the lowest class as 10−20.
(e) Convert the exclusive series constructed in (d) into 'less than' and 'more than' cumulative frequency distribution.

The two types of frequency distributions are inclusive and exclusive. State whether true or false.

Q. If class sizes are different in a given frequency distribution then to draw the histogram it is required to

Q. The time taken, in seconds, to solve a problem by each of 25 pupils is as follows:
16, 20, 26, 27, 28, 30, 33, 37, 38, 40, 42, 43, 46, 46, 46, 48, 49, 50, 53, 58, 59, 60, 64, 52, 20

(a) Construct a frequency distribution for these data, using a class interval of 10 seconds.
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40	45	38	24	46	42	45	18	53	64
45	32	52	54	78	65	52	64	66	43
48	55	50	43	48	20	27	65	37	55
51	55	62	66	38	16	60	58	46	35
72	62	54	58	30	36	43	82	46	53

40	45	38	24	46	42	45	18	53	64
45	32	52	54	78	65	52	64	66	43
48	55	50	43	48	20	27	65	37	55
51	55	62	66	38	16	60	58	46	35
72	62	54	58	30	36	43	82	46	53

In order to draw the histogram of the inclusive frequency distribution, it should be converted into exclusive form by applying a correction factor. A correction factor is calculated as

The correct option is A d2The given distribution is in inclusive form. It should be converted into exclusive form. This can be done by applying a correction factor d2, where d = (lower limit of a class) - (upper limit of a class before it)

The correct option is A $\frac{d}{2}$
The given distribution is in inclusive form. It should be converted into exclusive form. This can be done by applying a correction factor $\frac{d}{2}$ , where

d = (lower limit of a class) - (upper limit of a class before it)

40	45	38	24	46	42	45	18	53	64
45	32	52	54	78	65	52	64	66	43
48	55	50	43	48	20	27	65	37	55
51	55	62	66	38	16	60	58	46	35
72	62	54	58	30	36	43	82	46	53