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In Exercises 37鈥40, refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 11.5 years; (Exercise 38) 8.9 years; (Exercise 39) 59.5; (Exercise 40) 65.4.

Standard deviation for frequency distribution

s=nfx2-fx2nn-1

Blood Platelet Count of Males

Frequency

0-99

1

100-199

51

200-299

90

300-399

10

400-499

0

500-599

0

600-699

1

Short Answer

Expert verified

The standard deviation is computed as 68.4.

The value of 68.4 differs from the value of 59.5 by a significant amount. Therefore, these two values are not equal.

Step by step solution

01

Given information

The frequency distribution of the number of males corresponding to the blood platelet count is tabulated.

02

Formula

Thestandard deviationis a measure of dispersion most commonly used for a set of data. For a frequency distribution, the following formula is used:

s=nfx2-fx2nn-1

Here,

  • the frequencies are denoted by f;
  • the class midpoints are denoted by x;
  • the sum of frequencies is denoted by n.
03

Calculations

The calculations are tabulated below:

Blood Platelet Count

Frequency (f)

Midpoint

(x)

fx

fx2

0-99

1

49.5

49.5

2450.25

100-199

51

149.5

7624.5

1139863

200-299

90

249.5

22455

5602523

300-399

10

349.5

3495

1221503

400-499

0

449.5

0

0

500-599

0

549.5

0

0

600-699

1

649.5

649.5

421850.3


n=153

fx=34273.5

fx2=8388188

Substituting the above values in the formula:

s=nfx2-fx2nn-1=1538388188-34273.52153153-1=68.4

Therefore, the computed value of the standard deviation is68.4.

04

Compare the actual and computed value of standard deviation 

The actual value of standard deviation is 59.5.

The calculated value of the standard deviation equal to 68.4differs from the value obtained from the original list of observations, which is 59.5.

Thus, the two values cannot be considered equal.

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