Answered: PLEASE HELP ME TO GET THE HISTOGRAM,…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

PLEASE HELP ME TO GET THE HISTOGRAM, FREQUENCY POLYGON AND PIE CHART OF THE PICTURE BELOW.

THE SOLUTION WAS PROVIDED BELOW:

The mean gives the average of all the observation and it calculated for grouped frequency table as :

x¯=∑fx∑f

The median is the middle value of the data set and it calculated for grouped frequency table as :

Md=l+n2-cf×h

where n is the number of observation, l is the lowest boundary and f is the frequency and h is the class width.

Mode is the most repeated value of the data set and it calculated for grouped frequency table as :

Mo=l+f-f12f-f1-f2×h

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Solution

The data given for 40 students based on the 25 point test and it is defined as :

Scores	Class boundary	f	x	fx	fxi-x¯2	Cumulative frequency
5-7	4.5-7.5	2	6	12	136.125	2
8-10	7.5-10.5	4	9	36	110.25	6
11-13	10.5-13.5	14	12	168	70.875	20
14-16	13.5-16.5	9	15	135	5.0625	29
17-19	16.5-19.5	6	18	108	84.375	35
20-22	19.5-22.5	3	21	63	135.6875	38
23-25	22.5-25.5	2	24	48	190.125	40
Total		40		570	733.5

From the information given :

Percentile 20 :

Let, 20×n100=20×40100=8

The cumulative frequency greater than 8 is 20 and its corresponding frequency and lower boundary as 14 and 10.5 respectively. The class width is 3 and the cumulative value is 6.

Now,

P20=l+20n100-cf×h=10.5+8-614×3=10.5+0.429=10.929

Thus, P20=10.929

Decile 2 :

Let, 2×n10=2×4010=8

The cumulative frequency greater than 8 is 20 and its corresponding frequency and lower boundary as 14 and 10.5 respectively. The class width is 3 and the cumulative value is 6.

Now,

D2=l+2n10-cf×h=10.5+8-614×3=10.5+0.429=10.929

Thus, D2=10.929

Quartile 2 :

Let, 2×n4=2×404=20

The cumulative frequency greater than 20 is 29 and its corresponding frequency and lower boundary as 9 and 13.5 respectively. The class width is 3 and the cumulative value is 20.

Now,

Q2=l+2n4-cf×h=13.5+20-209×3=13.5+0=13.5

Thus, Q2=13.5

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Solution - 2

Using the table values :

Mean :

x¯=∑fx∑f=57040=14.25

Mean is 14.25

Median :

Let, n2=402=20

The cumulative frequency greater than 20 is 29 and its corresponding frequency and lower boundary as 9 and 13.5 respectively. The class width is 3 and the cumulative value is 20.

Now,

Md=l+n2-cf×h=13.5+20-209×3=13.5+0=13.5

Thus, Md=13.5

Mode :

Using the data given, highest frequency is 14 and its corresponding lower boundary as 10.5. The upper and lower frequency of highest frequency is 4 and 9 respectively. The class width is 3 .

Now,

Mo=l+f-f12f-f1-f2×h=10.5+14-42×14-4-9×3=10.5+2=12.5

Thus, Mo=12.5