Chapter 1.3, Problem 88E

(a)

To determine

To find: the median and the quartiles

Answer to Problem 88E

The median is 2, and the quartiles are 1 and 4 respectively

Explanation of Solution

Given:

  

Calculation:

From the given histogram of number of servings of fruit per day claimed by 74 seventeen years old girls is a study in Pennsylvania the Median is 2, first quartile is 1 and third quartile is 4. Calculate median and quartiles as follows:

Domain name length (X)	Frequency (f)	Cumulative frequency (Cf)
0	15	15
1	11	26
2	15	41
3	11	52
4	8	60
5	5	65
6	3	68
7	3	71
8	3	74

We can find the median as

  $\begin{array}{l}M=\frac{N+1}{2}\\ N=\text{Cumilative}\text{frequency}\\ M=\frac{74+1}{2}\\ M=37.5\text{th}\text{item}\\ =2\end{array}$

  $\begin{array}{l}{Q}_1=\frac{N+1}{4}\text{th}\text{item}\\ Q_1=\frac{75}{4}\text{th}\text{item}\\ Q_1=18.75\text{th}\text{item}\\ \text{=1}\end{array}$

  $\begin{array}{l}{Q}_3=\frac{3\left(N+1\right)}{4}\text{th}\text{item}\\ =\frac{3\left(75\right)}{4}\text{th}\text{item}\\ =56.25\text{th}\text{item}\\ \text{=4}\end{array}$

Conclusion:

Therefore, the median is 2, and the quartiles are 1 and 4 respectively

(b)

To determine

To estimate: the mean of the distribution.

Answer to Problem 88E

Mean is 2.59

Explanation of Solution

Calculation:

From the given histogram of number of servings of fruit per day claimed by 74 seventeen years old girls is a study in Pennsylvania the mean

Domain name length (X)	Frequency (f)	tx
0	15	0
1	11	11
2	15	30
3	11	33
4	8	32
5	5	25
6	3	18
7	3	21
8	3	24

The mean of the distribution is

  $\begin{array}{l}\overline{X}=\frac{{\displaystyle \sum fx}}{{\displaystyle \sum f}}\\ =\frac{194}{74}\\ \overline{X}=2.59\end{array}$

Conclusion:

Therefore, mean is 2.59