In Exercises 37–40, 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

$s=\sqrt{\frac{n\left[\sum \left(f\times x^2\right)\right]-{\left[\sum \left(f\times x\right)\right]}^2}{n\left(n-1\right)}}$

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

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

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

$s=\sqrt{\frac{n\left[\sum \left(f\times x^2\right)\right]-{\left[\sum \left(f\times x\right)\right]}^2}{n\left(n-1\right)}}$

Here,

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

The calculations are tabulated below:

Substituting the above values in the formula:

$$\begin{gathered} s=\sqrt{\frac{n\left[\sum \left(f\times x^2\right)\right]-{\left[\sum \left(f\times x\right)\right]}^2}{n\left(n-1\right)}} \\ =\sqrt{\frac{153\left(8388188\right)-{\left(34273.5\right)}^2}{153\left(153-1\right)}} \\ =68.4 \end{gathered}$$

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

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.