We appreciate your visit to Q3 From the following data find out the Standard Deviation and its coefficient of variation using the Step Deviation Method Age in years No of. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
To find the Standard Deviation and its coefficient of variation using the Step Deviation Method, follow these steps:
Step 1: Create a Frequency Table
For the given data of ages and number of students:
- Age 20-25: Frequency (f) = 170
- Age 25-30: Frequency (f) = 110
- Age 30-35: Frequency (f) = 80
- Age 35-40: Frequency (f) = 45
- Age 40-45: Frequency (f) = 40
- Age 45-50: Frequency (f) = 30
- Age 50-55: Frequency (f) = 25
Step 2: Determine the Midpoint (x) for Each Class Interval
- Midpoint for 20-25: [tex]\frac{20 + 25}{2} = 22.5[/tex]
- Midpoint for 25-30: [tex]\frac{25 + 30}{2} = 27.5[/tex]
- Midpoint for 30-35: [tex]\frac{30 + 35}{2} = 32.5[/tex]
- Midpoint for 35-40: [tex]\frac{35 + 40}{2} = 37.5[/tex]
- Midpoint for 40-45: [tex]\frac{40 + 45}{2} = 42.5[/tex]
- Midpoint for 45-50: [tex]\frac{45 + 50}{2} = 47.5[/tex]
- Midpoint for 50-55: [tex]\frac{50 + 55}{2} = 52.5[/tex]
Step 3: Choose an Assumed Mean (A)
Commonly, the midpoint of the class interval with the highest frequency is chosen as the assumed mean. Here, we choose 22.5.
Step 4: Calculate "d", the deviation from the assumed mean
- [tex]d = \frac{x - A}{c}[/tex], where [tex]c[/tex] is the class interval (5 in this case).
Calculations for d:
- d for 22.5: [tex]d = \frac{22.5 - 22.5}{5} = 0[/tex]
- d for 27.5: [tex]d = \frac{27.5 - 22.5}{5} = 1[/tex]
- d for 32.5: [tex]d = \frac{32.5 - 22.5}{5} = 2[/tex]
- d for 37.5: [tex]d = \frac{37.5 - 22.5}{5} = 3[/tex]
- d for 42.5: [tex]d = \frac{42.5 - 22.5}{5} = 4[/tex]
- d for 47.5: [tex]d = \frac{47.5 - 22.5}{5} = 5[/tex]
- d for 52.5: [tex]d = \frac{52.5 - 22.5}{5} = 6[/tex]
Step 5: Find [tex]fd[/tex] and [tex]f(d^2)[/tex]
- Calculate [tex]fd[/tex] and [tex]f(d^2)[/tex] for each class.
Age Interval
d
Frequency (f)
fd
f(d^2)
20-25
0
170
0
0
25-30
1
110
110
110
30-35
2
80
160
320
35-40
3
45
135
405
40-45
4
40
160
640
45-50
5
30
150
750
50-55
6
25
150
900
Sum of frequencies ([tex]N[/tex]) = 500
[tex]\sum fd = 865[/tex]
[tex]\sum f(d^2) = 3125[/tex]
Step 6: Calculate Mean and Standard Deviation
Mean, [tex]\bar{x} = A + (\frac{\sum fd}{N})\times c[/tex]
[tex]\bar{x} = 22.5 + (\frac{865}{500})\times 5 = 22.5 + 8.65 = 31.15[/tex]
Standard Deviation ([tex]\sigma[/tex]):
[tex]\sigma = c \times \sqrt{\frac{\sum f(d^2) - (\frac{(\sum fd)^2}{N})}{N}}[/tex]
[tex]\sigma = 5 \times \sqrt{\frac{3125 - (\frac{865^2}{500})}{500}}[/tex]
[tex]\sigma = 5 \times \sqrt{\frac{3125 - 1497.62}{500}}[/tex]
[tex]\sigma = 5 \times \sqrt{3.2556} = 5 \times 1.80434 = 9.02[/tex]
Step 7: Coefficient of Variation (CV)
[tex]CV = \frac{\sigma}{\bar{x}} \times 100\%[/tex]
[tex]CV = \frac{9.02}{31.15} \times 100\% = 28.96\%[/tex]
Thus, the Standard Deviation is 9.02 and the Coefficient of Variation is 28.96%.
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