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

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

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

Short answer

The standard deviation is 69.5.

The calculated value of the standard deviation is approximately equal to the actual value.

Step-by-step solution

Given information

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.

Formula for the standard deviation in the frequency distribution

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,

• f defines the frequencies;

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

Computation for the standard deviation

The terms of the formula are computed as shown below:

Blood Platelet Count	Frequency (f)	Midpoints (x)	$f\times x$	$f\times x^2$
100-199	25	49.5	1237.5	61256.25
200-299	92	149.5	13754	2056223
300-399	28	249.5	6986	1743007
400-499	0	349.5	0	0
500-599	2	449.5	899	404100.5
	n=147		$\sum \left(f\times x\right)=22876.5$	$\sum \left(f\times x^2\right)=4264587$

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.

Compare the computed and actual value of the standard deviation 

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.