Question

for the following data set calculate the percentage of data points that fall within one standard deviation of the mean, and compare the result to the expected percentage of a normal distribution. {50,46,54,51,29,52,48,54,47,48}

Answer 1

Answer:

First we need to find the mean and standard deviation of the given data.

The mean and standard deviation are given below:

$Mean=\frac{50+46+54+51+29+52+48+54+47+48}{10}=\frac{479}{10}=47.9$

$Standard-deviation=\sqrt{\frac{\sum(x-\bar{x})^2}{n-1} }$

                             $=\sqrt{\frac{(50-47.9)^2+(46-47.9)^2+(54-47.9)^2+...+(48-47.9)^2}{10-1} }$

                             $=7.20$

We have:

$\bar{x} \pm s=(47.9 \pm 7.20)$

                     $=(47.9-7.20, 47.9+7.20)$

                     $=(40.7, 55.1)$      

Therefore, the percentage of values that lies within one standard deviation of the mean is:

$\frac{9}{10} \times 100 =90\%$

The expected percentage of values within one standard deviation of the mean according to normal distribution is 68%.

Therefore, the observed percentage of values within one standard deviation of the mean is much higher than the expected percentage of a normal distribution.