Respuesta :
Answer:
First we need to find the mean and standard deviation of the given data.
The mean and standard deviation are given below:
[tex]Mean=\frac{50+46+54+51+29+52+48+54+47+48}{10}=\frac{479}{10}=47.9[/tex]
[tex]Standard-deviation=\sqrt{\frac{\sum(x-\bar{x})^2}{n-1} }[/tex]
[tex]=\sqrt{\frac{(50-47.9)^2+(46-47.9)^2+(54-47.9)^2+...+(48-47.9)^2}{10-1} }[/tex]
[tex]=7.20[/tex]
We have:
[tex]\bar{x} \pm s=(47.9 \pm 7.20)[/tex]
[tex]=(47.9-7.20, 47.9+7.20)[/tex]
[tex]=(40.7, 55.1)[/tex]
Therefore, the percentage of values that lies within one standard deviation of the mean is:
[tex]\frac{9}{10} \times 100 =90\%[/tex]
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.
