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

Example: Grouped Data
Compute the variance and standard deviation of the following grade
distribution of 40 students based on their 25-point test.
Step 2.
fx
f(x – x)²
Scores
f
Solve for Variance
|
5-7
2
6
12
136.1250
Ef(x-x)2
n - 1
s2
8-10
4
9.
36
110.2500
733.5
11-13
14
12
168
70.8750
s2
40 – 1
14-16
15
135
5.0625
s2
18.8077
%3D
17-19
18
108
84.3750
Step 3.
20-22
3
21
63
135.6875
Solve for Standard
Deviation
23-25
2
24
48
190.125
s = 4.34
n=40
570
733.5

Definition Definition Measure of central tendency that is the value that occurs most frequently in a data set. A data set may have more than one mode if multiple categories repeat an equal number of times. For example, in a data set with five item—3, 5, 5, 29, 473—the mode is 5 because it occurs twice and no other value occurs more than once. On a histogram or bar chart, the element with the highest bar represents the mode. Therefore, the mode is sometimes considered the most popular option. The mode is useful for nominal or categorical data (e.g., the most common color car that users purchase), but it is problematic for continuous data because it is more likely not to have any value that is more frequent than the other.

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