Question

Examine the sample data in the accompanying table.

5.9 5.3 1.6 7.4 8.6 1.2 2.1

4.0 7.3 8.4 8.9 6.7 4.5 6.3

7.6 9.7 3.5 1.1 4.3 3.3 8.4

1.6 8.2 6.5 1.1 5.0 9.4 6.4

a. Construct a stem-and-leaf plot to assess whether thedata are from an approximately normal distribution.

b. Compute sfor the sample data.

c. Find the values of Q_L and Q_U, then use these values andthe value of sfrom part b to assess whether the data comefrom an approximately normaldistribution.

d. Generate a normal probability plot for the data and useit to assess whether the data are approximately normal.

Short answer

a. The plot is not symmetric. So, the data is not normal.

b. The standard deviation is 2.765172

c. The $Q_L=3.4$, and $Q_U=7.9$. the data is not normal.

d. The normal probability plot shows that the data is not normal

Step-by-step solution

Given information

Here sample data is provided.

Stem and leaf plot

a.

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.

Calculating standard deviation

b.

No	x	${\left(x_i-\overrightarrow{x}\right)}^2$
1	5.9	0.151543
2	5.3	0.044401
3	1.6	15.29369
4	7.4	3.569401
5	8.6	9.543686
6	1.2	18.58226
7	2.1	11.63297
8	4	2.282258
9	7.3	3.201543
10	8.4	8.347972
11	8.9	11.48726
12	6.7	1.414401
13	4.5	1.021543
14	6.3	0.622972
15	7.6	4.365115
16	9.7	17.55011
17	3.5	4.042972
18	1.1	19.4544
19	4.3	1.465829
20	3.3	4.887258
21	8.4	8.347972
22	1.6	15.29369
23	8.2	7.232258
24	6.5	0.978686
25	1.1	19.4544
26	5	0.260829
27	9.4	15.12654
28	6.4	0.790829
Total	154.3	206.4468

The mean is calculated as follows :

$$\begin{gathered} \overrightarrow{x}=\frac{{\displaystyle \sum _{i=1}^n}{x}_i}{n} \\ =\frac{154.3}{28} \\ =5.510714 \end{gathered}$$

Thus, the mean is 5.510714.

The standard deviation is calculated as follows :

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

Thus, the standard deviation is 2.765172.

Checking for normality

No		observation
1		1.1
2		1.1
3		1.2
4		1.6
5		1.6
6		2.1
7		3.3
8		3.5
9		4
10		4.3
11		4.5
12		5
13		5.3
14		5.9
15		6.3
16		6.4
17		6.5
18		6.7
19		7.3
20		7.4
21		7.6
22		8.2
23		8.4
24		8.4
25		8.6
26		8.9
27		9.4
28		9.7

The lower quantile Q_L is,

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

Thus,the lower quantile is 3.4

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

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.

Checking for normality using graph

d.

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