Question

for the following set of data, find the percentage of data within 1 population standard deviation of the mean, to the nearest 10th of a percent.

data	frequency
18	7
30	11
44	14
47	11
53	8
58	1

Explanation:

Step1: Calculate the mean

First, find the sum of the products of data - values and their frequencies. Let $x_i$ be the data - values and $f_i$ be the frequencies. The total number of data points $n=\sum_{i = 1}^{k}f_i$, where $k$ is the number of data - value groups.
$n=2 + 7+11 + 14+11+8+1=54$
$\sum_{i = 1}^{7}x_if_i=14\times2 + 18\times7+30\times11+44\times14+47\times11+53\times8+58\times1$
$=28+126 + 330+616+517+424+58$
$=2099$
The mean $\mu=\frac{\sum_{i = 1}^{7}x_if_i}{n}=\frac{2099}{54}\approx38.87$

Step2: Calculate the population variance

The population variance $\sigma^{2}=\frac{\sum_{i = 1}^{k}f_i(x_i-\mu)^{2}}{n}$
$(14 - 38.87)^{2}\times2+(18 - 38.87)^{2}\times7+(30 - 38.87)^{2}\times11+(44 - 38.87)^{2}\times14+(47 - 38.87)^{2}\times11+(53 - 38.87)^{2}\times8+(58 - 38.87)^{2}\times1$
$=(- 24.87)^{2}\times2+(-20.87)^{2}\times7+(-8.87)^{2}\times11+(5.13)^{2}\times14+(8.13)^{2}\times11+(14.13)^{2}\times8+(19.13)^{2}\times1$
$=618.5169\times2 + 435.5569\times7+78.6769\times11+26.3169\times14+66.0969\times11+199.6569\times8+365.9569\times1$
$=1237.0338+3048.8983+865.4459+368.4366+727.0659+1597.2552+365.9569$
$=8210.1926$
$\sigma^{2}=\frac{8210.1926}{54}\approx152.04$
The population standard deviation $\sigma=\sqrt{152.04}\approx12.33$

Step3: Find the range within 1 standard - deviation of the mean

The lower limit is $L=\mu-\sigma\approx38.87 - 12.33 = 26.54$
The upper limit is $U=\mu+\sigma\approx38.87+12.33 = 51.2$

Step4: Count the number of data points within the range

For $x = 30$, $f = 11$; for $x = 44$, $f = 14$; for $x = 47$, $f = 11$
The number of data points within the range is $11 + 14+11=36$

Step5: Calculate the percentage

The percentage of data within 1 population standard deviation of the mean is $\frac{36}{54}\times100\%\approx66.7\%$

Answer:

$66.7\%$