91影视

Here sample data is provided.

The decimal point is at the |

1 | 11266

2 | 1

3 | 35

4 | 035

5 | 039

6 | 3457

7 | 346

8 | 24469

9 | 47

The plot is not symmetric.

So, the data is not normal.

b.

The mean is calculated as follows :

$$\begin{gathered} \stackrel{鈫�}{x}=\frac{{\displaystyle \underset{i=1}{\overset{n}{鈭�}}}{x}_i}{n} \\ =\frac{154.3}{28} \\ =5.510714 \end{gathered}$$

Thus, the mean is 5.510714.

The standard deviation is calculated as follows :

$$\begin{gathered} s=\sqrt{\frac{{\displaystyle \underset{i=1}{\overset{n}{鈭�}}{\left(x_i-\overline{x}\right)}^2}}{n-1}} \\ =\sqrt{\frac{206.4468}{28-1}} \\ =2.765172 \end{gathered}$$

Thus, the standard deviation is 2.765172.

The lower quantile Q_L is,

$$\begin{gathered} Q_L=\frac{1}{4}{\left(n+1\right)}^{th}observation \\ =0.25脳{\left(28+1\right)}^{th}observation \\ =7.{25}^{th}observation \\ =\frac{7^{th}+8^{th}}{2}observation \\ =\frac{3.3+3.5}{2} \\ =3.4 \end{gathered}$$

Thus,the lower quantile is 3.4

The upper quantile is,$$\begin{gathered} Q_U=\frac{3}{4}{\left(n+1\right)}^{th}observation \\ =0.75脳{\left(28+1\right)}^{th}observation \\ =21.{75}^{th}observation \\ =\frac{{21}^{th}+{22}^{th}}{2}observation \\ =\frac{7.6+8.2}{2} \\ =7.9 \end{gathered}$$

Thus, the upper quantile is 7.9.

$$\begin{gathered} \frac{IQR}{s}=\frac{Q_U-Q_L}{s} \\ =\frac{7.9-3.4}{2.765172} \\ =1.627385 \end{gathered}$$

Here, the value of IQR/s is not approximately equal to 1.3

So, the data is not normal.

d.

The normal probability plot shows that all data points are not in line. So, the data points are not normally distributed.