Question

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

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

Age (yr) of Best Actress When Oscar Was Won	Frequency
20-29	29
30-39	34
40-49	14
50-59	3
60-69	5
70-79	1
80-89	1

Short answer

The standard deviation for the given frequency distribution is equal to 12.69 years.

The value of 12.69 years is close to the value of 11.5 years.

Step-by-step solution

Given information

A frequency distribution table is given, showing the frequencies of the ages of the best actresses who won the Oscar at that time.

The table consists of 7 class intervals.

The actual standard deviation value is 11.5 years.

Formula for the standard deviation of frequency distribution

Thestandard deviation of a frequency distribution is computed using the given formula:

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

Here, f represents the frequencies;

x represents the midpoints of the class intervals;

n represents the total frequency.

Calculations for computing the standard deviation values

The computations are done using the table as shown below:

Age (in years)	Midpoint (x)	Frequency (f)	$f\times x$	$x^2$	$f\times x^2$
20-29	24.5	29	710.5	600.25	17407.25
30-39	34.5	34	1173	1190.25	40468.5
40-49	44.5	14	623	1980.25	27723.5
50-59	54.5	3	163.5	2970.25	8910.75
60-69	64.5	5	322.5	4160.25	20801.25
70-79	74.5	1	74.5	5550.25	5550.25
80-89	84.5	1	84.5	7140.25	7140.25

Therefore, the following values are computed as follows:

$$\begin{gathered} \sum \left(f\times x\right)=3151.5 \\ {\left[\sum \left(f\times x\right)\right]}^2=9931952.25 \\ \sum \left(f\times x^2\right)=128001.8 \end{gathered}$$

The total count of observation n is $\sum f=87$.

Thus, the value of standard deviation is as given below:

$$\begin{gathered} s=\sqrt{\frac{n\left[\sum \left(f\times x^2\right)\right]-{\left[\sum \left(f\times x\right)\right]}^2}{n\left(n-1\right)}} \\ =\sqrt{\frac{87\left(128001.8\right)-{\left(3151.5\right)}^2}{87\left(87-1\right)}} \\ =12.69 \end{gathered}$$

The computed value of the standard deviation is equal to12.69 years.

Compare the computed and actual values

The actual value of standard deviation is 11.5 years.

The calculated value equal to 12.69 years isclose to the actual mean value of 11.5 years.