91Ó°ÊÓ

Consider

$\begin{array}{l}\stackrel{¯}{x}=4.71\\ s=6.09\end{array}$

$\begin{array}{l}{Q}_L=1\\ Q_U=6\end{array}$

General conditions:

1. If the mean>median, the distribution is right skewed
2. If the mean<median, the distribution is left skewed.

If the mean=median, then it is normal distribution

The median is the half way between the lower and upper quartile.

$\begin{array}{c}Median=\frac{Q_1+Q_2}{2}\\ =\frac{1+6}{2}\\ =\frac{7}{2}\end{array}$

$Median=3.5$

Hence the value of the median is 3.5.

Here the mean is larger than median.

That is, $mean\left(\stackrel{¯}{x}\right)=4.71>median=3.5$

This shows that the distribution is right skewed. Hence the data is not normally distributed.

Calculate the inter quartile range and the standard deviation to obtain the ratio $\frac{IQR}{S}$

If the data is approximately normal then the ratio$\frac{IQR}{S}≈1.3$

The ratio

$\begin{array}{c}\frac{IQR}{S}=\frac{Q_U−Q_L}{S}\\ =\frac{6−1}{6.09}\\ =\frac{5}{6.09}\end{array}$

$\frac{IQR}{S}=0.82$

Hence the value of ratio is 0.82.

Here the ratio is very small when compared to the value1.3 this clearly indicates that the data is not approximately normally distributed.

Here the standard deviation is greater than the mean.

That is $\stackrel{¯}{x}=4.71<SD=6.09$

This indicates that the data is skewed to right. Therefore, the data is not normal distribution.