Chapter 13, Problem 2PS

a)

To determine

To graph: A stem and leaf plot of given data is to be drawn.

Answer to Problem 2PS

Median $=35$

Range $=46$

Explanation of Solution

Given information: Data of ages of adult male passengers on the Mayflower at the time of its departure

  $\begin{array}{l}21,34,29,38,30,54,39,20,35,64,37,45,21,\hfill \\ 25,55,45,40,38,38,21,21,34,38,52,41,48,\hfill \\ 18,27,32,49,30,42,30,25,38,25,20\hfill \end{array}$

Graph:

  

Key: $1|8=18$

Interpretation:

Leaves represent the unit digits of the numbers and stem represent the tens digits.

b)

To determine

The median and the range of the ages are to be calculated.

Answer to Problem 2PS

Median $=35$

Range $=46$

Explanation of Solution

Given: Data of ages of adult male passengers on the Mayflower at the time of its departure

  $\begin{array}{l}21,34,29,38,30,54,39,20,35,64,37,45,21,\hfill \\ 25,55,45,40,38,38,21,21,34,38,52,41,48,\hfill \\ 18,27,32,49,30,42,30,25,38,25,20\hfill \end{array}$

Concept Used: Formula of median:

If the number of observations (n) is odd, then median is $=\left(\frac{n+1}{2}\right)$

Range $=$ the maximum value $-$ the minimum value

Calculation: For median:

Firstly, arrange the data in increasing order

  $\begin{array}{l}\text{18, 20, 20, 21, 21, 21, 21, 25, 25, 25,}\\ \text{27, 29, 30, 30, 30, 32, 34, 34, 35, 37,}\\ \text{38, 38, 38, 38, 38, 39, 40, 41, 42, 45, }\\ \text{45, 48, 49, 52, 54, 55, 64}\end{array}$

Here, number of observations is odd $\left(n=37\right)$

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

  $=\left(\frac{37+1}{2}\right)$

  $=\frac{38}{2}$

  $=16$

So, the 16^th observation is $=35$

Range $=$ the maximum value $-$ the minimum value

Range $=64-18$

  $=46$

Conclusion:

Median $=35$

Range $=46$

b)

To determine

The median and the range of the ages are to be calculated.

Answer to Problem 2PS

Median $=35$

Range $=46$

Explanation of Solution

Given: Data of ages of adult male passengers on the Mayflower at the time of its departure

  $\begin{array}{l}21,34,29,38,30,54,39,20,35,64,37,45,21,\hfill \\ 25,55,45,40,38,38,21,21,34,38,52,41,48,\hfill \\ 18,27,32,49,30,42,30,25,38,25,20\hfill \end{array}$

Concept Used: Formula of median:

If the number of observations (n) is odd, then median is $=\left(\frac{n+1}{2}\right)$

Range $=$ the maximum value $-$ the minimum value

Calculation: For median:

Firstly, arrange the data in increasing order

  $\begin{array}{l}\text{18, 20, 20, 21, 21, 21, 21, 25, 25, 25,}\\ \text{27, 29, 30, 30, 30, 32, 34, 34, 35, 37,}\\ \text{38, 38, 38, 38, 38, 39, 40, 41, 42, 45, }\\ \text{45, 48, 49, 52, 54, 55, 64}\end{array}$

Here, number of observations is odd $\left(n=37\right)$

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

  $=\left(\frac{37+1}{2}\right)$

  $=\frac{38}{2}$

  $=16$

So, the 16^th observation is $=35$

Range $=$ the maximum value $-$ the minimum value

Range $=64-18$

  $=46$

Conclusion:

Median $=35$

Range $=46$

c)

To determine

To find the probability of age of Thomas English passenger that having age of $18-29$ years

Answer to Problem 2PS

Probability is $0.4$

Explanation of Solution

Given: Data of ages of adult male passengers on the Mayflower at the time of its departure

  $\begin{array}{l}21,34,29,38,30,54,39,20,35,64,37,45,21,\hfill \\ 25,55,45,40,38,38,21,21,34,38,52,41,48,\hfill \\ 18,27,32,49,30,42,30,25,38,25,20\hfill \end{array}$

Concept Used: Probability P(A) $=$ $\frac{n(E)}{n(S)}$

Where, P (A) is the Probability of an event

  $n(E)$ is the number of favorable outcomes

  $n(S)$ is the total number of events

Calculation: Probability P (A) $=$ $\frac{n(E)}{n(S)}$

  $n(E)$ = 12, $n(S)$ = 37

P (A) $=$ $\frac{12}{37}$ = 0.4

  $n(E)$ is number of outcomes having age between $18-29$ years.

Conclusion: Probability is $0.4$

d)

To determine

To graph: A cumulative frequency distribution and cumulative frequency histogram for the data is to be drawn.

Explanation of Solution

Given information: Data of ages of adult male passengers on the Mayflower at the time of its departure

  $\begin{array}{l}21,34,29,38,30,54,39,20,35,64,37,45,21,\hfill \\ 25,55,45,40,38,38,21,21,34,38,52,41,48,\hfill \\ 18,27,32,49,30,42,30,25,38,25,20\hfill \end{array}$

Graph: Cumulative frequency distribution

Interval	Frequency	Cumulative Frequency
[15,25)	7	7
[25,35)	11	7+11=18
[35,45)	11	7+11+11=29
[45,55)	6	7+11+11+6=35
[55,65)	2	7+11+11+6+2=37

Cumulative frequency histogram:

  

Interpretation: Cumulative frequency is the running total of frequencies.