Chapter 4, Problem 30E

(a)

To determine

To find: the median and IQR.

Answer to Problem 30E

Median and IQR are 66 and 5 respectively.

Explanation of Solution

Given:

Height	Count	Height	Count
60	2	69	5
61	6	70	11
62	9	71	8
63	7	72	9
64	5	73	4
65	20	74	2
66	18	75	4
67	7	76	1
68	12

Formula used:

For the even number

  $Median=\frac{\left(\frac{n}{2}\right)thvalue+\left(\frac{n}{2}+1\right)thvalue}{2}$

  $IQR=Q_3-Q_1$

Calculation:

  $\begin{array}{l}Median=\frac{\left(\frac{n}{2}\right)thvalue+\left(\frac{n}{2}+1\right)thvalue}{2}\\ =\frac{\left(\frac{130}{2}\right)thvalue+\left(\frac{130}{2}+1\right)thvalue}{2}\end{array}$

  $=\frac{65thvalue+66thvalue}{2}$

  $\begin{array}{l}=\frac{66+66}{2}\\ =66\end{array}$

Median is 66 inches

Every half of the data is having 65 values. Lower quartile is value at the 33^rd position in lower half and upper quartile is value would be at 33^rd position at upper half therefore

  $\begin{array}{l}{Q}_1=65\\ Q_3=70\end{array}$

  $\begin{array}{c}IQR=Q_3-Q_1\\ =70-65\\ =5\end{array}$

Therefore the IQR would be 5

(b)

To determine

To find: the mean and standard deviation.

Answer to Problem 30E

Mean and standard deviation are 67 and 3.8 respectively.

Explanation of Solution

Given:

Height	Count	Height	Count
60	2	69	5
61	6	70	11
62	9	71	8
63	7	72	9
64	5	73	4
65	20	74	2
66	18	75	4
67	7	76	1
68	12

Formula used:

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

  $\sigma =\sqrt{\frac{f{\left(mean-h\right)}^2}{n-1}}$

Calculation:

Height (h)	Count (f)	f*h	${\left(mean-h\right)}^2$	f* ${\left(mean-h\right)}^2$
60	2	120	49	98
61	6	366	36	216
62	9	558	25	225
63	7	441	16	112
64	5	320	9	45
65	20	1300	4	80
66	18	1188	1	18
67	7	469	0	0
68	12	816	1	12
69	5	345	4	20
70	11	770	9	99
71	8	568	16	128
72	9	648	25	225
73	4	292	36	144
74	2	148	49	98
75	4	300	64	256
76	1	76	81	81
Total	130	8725		1857

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{8725}{130}\\ =67inches\end{array}$

For the standard deviation

  $\begin{array}{c}\sigma =\sqrt{\frac{f{\left(mean-h\right)}^2}{n-1}}\\ =\sqrt{\frac{1857}{130-1}}\\ =3.8\end{array}$

Therefore the mean is 67 and standard deviation is 3.8

(c)

To determine

To construct: the histogram on the basis of given data.

Answer to Problem 30E

Skewed to right

Explanation of Solution

Given:

Height	Count	Height	Count
60	2	69	5
61	6	70	11
62	9	71	8
63	7	72	9
64	5	73	4
65	20	74	2
66	18	75	4
67	7	76	1
68	12

Graph:

Histogram on the basis of given data

  

By seeing the histogram, it is observed that skewed to right because the right tail is longer than the left.

(d)

To determine

To Explain: the distribution.

Explanation of Solution

The distribution is slightly skewed right the reason is that the right tail is longer than the left. Because median and IQR are unbiased whereas centre and standard deviation are they should be used as the measures for centre and spread instead in this case.