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 Females	Frequency
100-199	25
200-299	92
300-399	28
400-499	0
500-599	2

The table given displays the number of females for each interval of blood platelet count.

There are five intervals of blood platelet count.

The actual value for standard deviation is 65.4.

For a grouped frequency distribution, thestandard deviation is computed using the following expression:

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

Here,

• x defines the midpoints of the class intervals;
• n defines the total of all the frequencies.

The terms of the formula are computed as shown below:

Substituting the above values in the formula:

$$\begin{gathered} s=\sqrt{\frac{n\left[\sum \left(f\times x^2\right)\right]-{\left[\sum \left(f\times x\right)\right]}^2}{n\left(n-1\right)}} \\ =\sqrt{\frac{147\left(4264587\right)-{\left(22876.5\right)}^2}{147\left(147-1\right)}} \\ =69.5 \end{gathered}$$

Therefore, the calculated standard deviation is equal to69.5.

The actual value is 65.4.

The standard deviation value computed from the sample and the original list of values isapproximately equal to each other.