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for the following set of data, find the population standard deviation, …

Question

for the following set of data, find the population standard deviation, to the nearest hundredth.

datafrequency
149
286
294
306
326
396
403
502

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Explanation:

Step1: Calculate the mean $\mu$

First, find the sum of the products of data - value and frequency ($\sum_{i = 1}^{n}x_if_i$) and the total frequency ($N=\sum_{i = 1}^{n}f_i$).
$\sum_{i = 1}^{n}x_if_i=13\times7 + 14\times9+28\times6 + 29\times4+30\times6+32\times6+39\times6+40\times3+50\times2$
$=91+126 + 168+116+180+192+234+120+100$
$=1327$
$N=7 + 9+6+4+6+6+6+3+2=53$
$\mu=\frac{\sum_{i = 1}^{n}x_if_i}{N}=\frac{1327}{53}\approx25.04$

Step2: Calculate the sum of squared - deviations $\sum_{i = 1}^{n}f_i(x_i-\mu)^2$

For $x_1 = 13,f_1 = 7$: $f_1(x_1-\mu)^2=7\times(13 - 25.04)^2=7\times(- 12.04)^2=7\times144.9616 = 1014.7312$
For $x_2 = 14,f_2 = 9$: $f_2(x_2-\mu)^2=9\times(14 - 25.04)^2=9\times(-11.04)^2=9\times121.8816 = 1096.9344$
For $x_3 = 28,f_3 = 6$: $f_3(x_3-\mu)^2=6\times(28 - 25.04)^2=6\times(2.96)^2=6\times8.7616 = 52.5696$
For $x_4 = 29,f_4 = 4$: $f_4(x_4-\mu)^2=4\times(29 - 25.04)^2=4\times(3.96)^2=4\times15.6816 = 62.7264$
For $x_5 = 30,f_5 = 6$: $f_5(x_5-\mu)^2=6\times(30 - 25.04)^2=6\times(4.96)^2=6\times24.6016 = 147.6096$
For $x_6 = 32,f_6 = 6$: $f_6(x_6-\mu)^2=6\times(32 - 25.04)^2=6\times(6.96)^2=6\times48.4416 = 290.6496$
For $x_7 = 39,f_7 = 6$: $f_7(x_7-\mu)^2=6\times(39 - 25.04)^2=6\times(13.96)^2=6\times194.8816 = 1169.2896$
For $x_8 = 40,f_8 = 3$: $f_8(x_8-\mu)^2=3\times(40 - 25.04)^2=3\times(14.96)^2=3\times223.8016 = 671.4048$
For $x_9 = 50,f_9 = 2$: $f_9(x_9-\mu)^2=2\times(50 - 25.04)^2=2\times(24.96)^2=2\times623.0016 = 1246.0032$
$\sum_{i = 1}^{n}f_i(x_i-\mu)^2=1014.7312+1096.9344+52.5696+62.7264+147.6096+290.6496+1169.2896+671.4048+1246.0032 = 5752.9184$

Step3: Calculate the population standard deviation $\sigma$

The formula for the population standard deviation of a frequency - distribution is $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\mu)^2}{N}}$
$\sigma=\sqrt{\frac{5752.9184}{53}}\approx\sqrt{108.5456}\approx10.42$

Answer:

$10.42$