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

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

Age (yr) of Best Actor When Oscar Was Won	Frequency
20-29	1
30-39	28
40-49	36
50-59	15
60-69	6
70-79	1

The data showed the frequencies of the best actors according to their ages when they won the Oscar. The ages are grouped into 7 class intervals.

The formula for calculating thestandard deviation of a frequency distribution is expressed as follows:

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

Here, f denotes the frequencies;

x denotes the midpoints of the class intervals;

n denotes the total frequency.

The table below shows the necessary calculations:

The value of standard deviation is calculated as follows, using the values obtained above:

$$\begin{gathered} s=\sqrt{\frac{n\left[\sum \left(f\times x^2\right)\right]-{\left[\sum \left(f\times x\right)\right]}^2}{n\left(n-1\right)}} \\ =\sqrt{\frac{87\left(180281.75\right)-{\left(3871.5\right)}^2}{87\left(87-1\right)}} \\ =9.6 \end{gathered}$$

Therefore, the computed standard deviation is equal to9.6 years.

The actual value of standard deviation is 8.9 years.

Thus, the standard deviation computed from the sample is 9.6 years, which isquite close to the actual value equal to 8.9 years.