Approximate the standard deviation for the following grouped frequency distribution table (GFDT):
| Data | Frequency |
| :------ | :-------- |
| 40-44 | 1 |
| 45-49 | 1 |
| 50-54 | 2 |
| 55-59 | 7 |
| 60-64 | 5 |
| 65-69 | 7 |
| 70-74 | 15 |
| 75-79 | 24 |
| 80-84 | 10 |
Sample standard deviation = $\square$
Enter an integer or decimal number. Round your answer to one decimal place.
Answer (1)
Calculate the midpoint of each class interval.
Calculate the weighted mean: x ˉ ≈ 71.0972 .
Calculate the sample standard deviation: s ≈ 9.279 .
Round the result to one decimal place: 9.3 .
Explanation
Understand the problem We are given a grouped frequency distribution table (GFDT) and asked to approximate the sample standard deviation. The data is grouped into intervals, and we have the corresponding frequencies for each interval. We need to find the sample standard deviation and round it to one decimal place.
Calculate midpoints First, we need to calculate the midpoint of each class interval. The midpoint m i is calculated as the average of the lower and upper limits of the interval: m i = 2 lower limit + upper limit .
For the given data:
40-44: m 1 = 2 40 + 44 = 42
45-49: m 2 = 2 45 + 49 = 47
50-54: m 3 = 2 50 + 54 = 52
55-59: m 4 = 2 55 + 59 = 57
60-64: m 5 = 2 60 + 64 = 62
65-69: m 6 = 2 65 + 69 = 67
70-74: m 7 = 2 70 + 74 = 72
75-79: m 8 = 2 75 + 79 = 77
80-84: m 9 = 2 80 + 84 = 82
Calculate weighted mean Next, we calculate the weighted mean x ˉ using the formula: x ˉ = ∑ f i ∑ f i m i , where f i is the frequency of the i -th class.
∑ f i = 1 + 1 + 2 + 7 + 5 + 7 + 15 + 24 + 10 = 72 ∑ f i m i = ( 1 ) ( 42 ) + ( 1 ) ( 47 ) + ( 2 ) ( 52 ) + ( 7 ) ( 57 ) + ( 5 ) ( 62 ) + ( 7 ) ( 67 ) + ( 15 ) ( 72 ) + ( 24 ) ( 77 ) + ( 10 ) ( 82 ) = 42 + 47 + 104 + 399 + 310 + 469 + 1080 + 1848 + 820 = 5119 x ˉ = 72 5119 ≈ 71.0972
Calculate sample standard deviation Now, we calculate the sample standard deviation s using the formula: s = n − 1 ∑ f i ( m i − x ˉ ) 2 , where n = ∑ f i is the total frequency.
∑ f i ( m i − x ˉ ) 2 = 1 ( 42 − 71.0972 ) 2 + 1 ( 47 − 71.0972 ) 2 + 2 ( 52 − 71.0972 ) 2 + 7 ( 57 − 71.0972 ) 2 + 5 ( 62 − 71.0972 ) 2 + 7 ( 67 − 71.0972 ) 2 + 15 ( 72 − 71.0972 ) 2 + 24 ( 77 − 71.0972 ) 2 + 10 ( 82 − 71.0972 ) 2 ≈ 1 ( 846.73 ) + 1 ( 580.73 ) + 2 ( 364.73 ) + 7 ( 198.73 ) + 5 ( 82.73 ) + 7 ( 16.78 ) + 15 ( 0.81 ) + 24 ( 34.81 ) + 10 ( 118.81 ) ≈ 846.73 + 580.73 + 729.46 + 1391.11 + 413.65 + 117.46 + 12.15 + 835.44 + 1188.1 ≈ 6114.83
s = 72 − 1 6114.83 = 71 6114.83 ≈ 86.1244 ≈ 9.279
Round the result Finally, we round the calculated standard deviation to one decimal place: s ≈ 9.3
State the final answer The approximate sample standard deviation for the given grouped frequency distribution is 9.3 .
Examples
Understanding standard deviation is crucial in many real-world scenarios. For instance, in quality control, manufacturers use standard deviation to ensure the consistency of their products. If the standard deviation of a product's dimensions is too high, it indicates that the product is not being manufactured consistently, and adjustments need to be made. Similarly, in finance, standard deviation is used to measure the volatility of investments. A high standard deviation indicates that the investment is more risky, while a low standard deviation indicates that the investment is more stable. By understanding and calculating standard deviation, we can make more informed decisions in various aspects of life.
