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

Transcribed Image Text: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

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

arrow_forward

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

Accepted Answer

Each variable can be numerical or categorical. Numerical variables are represented using the graphs…