185 Software file updates. Software configuration management was used to monitor a software engineering team’s performance at Motorola, Inc. (Software Quality Professional, Nov. 2004). One of the variables of interest was the number of updates to a file that was changed because of a problem report. Summary statistics for$\mathit{n}\mathbf{=}\mathbf{421}$ n = 421 files yielded the following results: role="math" localid="1658219642985" $\overline{\mathbf{x}}\mathbf{=}\mathbf{4}\mathbf{.71}$,$\mathit{s}\mathbf{=}\mathbf{6}\mathbf{.09}$, ${\mathit{Q}}_{\mathbf{L}}\mathbf{=}\mathbf{1}$, and${\mathit{Q}}_{\mathbf{U}}\mathbf{=}\mathbf{6}$ . Are these data approximately normally distributed? Explain.

Consider

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

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

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, $mean\left(\overline{x}\right)=4.71>median=3.5$

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$\frac{IQR}{S}\approx 1.3$

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 $\overline{x}=4.71<SD=6.09$

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