Question

question for the following set of data, find the percentage of data within 2 population standard deviations of the mean, to the nearest 10th of a percent. 55, 57, 61, 60, 61, 6, 5

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{55 + 57+61+60+61+6+5}{7}=\frac{305}{7}\approx43.57$

Step2: Calculate the population - variance

First, find the squared - differences from the mean:
$(55 - 43.57)^2\approx131.65$, $(57 - 43.57)^2\approx180.36$, $(61 - 43.57)^2\approx303.80$, $(60 - 43.57)^2\approx270.94$, $(61 - 43.57)^2\approx303.80$, $(6 - 43.57)^2\approx1411.65$, $(5 - 43.57)^2\approx1487.65$
The population variance $\sigma^{2}=\frac{131.65 + 180.36+303.80+270.94+303.80+1411.65+1487.65}{7}=\frac{4089.85}{7}\approx584.26$

Step3: Calculate the population - standard deviation

$\sigma=\sqrt{584.26}\approx24.17$

Step4: Find the range within 2 standard deviations of the mean

The lower bound is $\bar{x}-2\sigma=43.57-2\times24.17=43.57 - 48.34=-4.77$
The upper bound is $\bar{x}+2\sigma=43.57 + 48.34=91.91$

Step5: Count the number of data points within the range

The data points $55,57,61,60,61$ are within the range. There are 5 data points.
The total number of data points is $n = 7$.
The percentage is $\frac{5}{7}\times100\%\approx71.4\%$

Answer:

$71.4\%$