Chapter 6, Problem 4RE

a.

To determine

To Construct the probability distribution of the given data.

Explanation of Solution

Given:

Number of Tests	Frequency (in 1000s)
1	953
2	423
3	194
4	80
5	29
6	9
7	3
8	1
Total	1692

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

b.

To determine

To find the probability that student took only one exam.

Answer to Problem 4RE

Probability that student took only one exam = 0.563

Explanation of Solution

Calculation:

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Calculation:

  $\begin{array}{c}P\left(\text{That}\text{student took only one exam}\right)=\frac{953}{1692}\\ =0.563\end{array}$

c.

To determine

To find the mean ${\mu }_x$

Answer to Problem 4RE

The mean of the given distribution is 1.73.

Explanation of Solution

Formula Used:

  ${\mu }_x=\sum x.p\left(x\right)$

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	$x.P\left(x\right)$
1	953	0.563	0.56
2	423	0.250	0.50
3	194	0.115	0.34
4	80	0.047	0.19
5	29	0.017	0.09
6	9	0.005	0.03
7	3	0.002	0.01
8	1	0.001	0.00
Total	1692	1.000	1.73

The mean ${\mu }_x$ of a given distribution is 1.73, because the value of $\sum x.p\left(x\right)$ = 1.73. Therefore, the mean is 1.73.

d.

To determine

To find the mean ${\mu }_x$

Answer to Problem 4RE

Standard Deviation = 1.05

Explanation of Solution

Formula Used:

  ${\mu }_x=\sum x.p\left(x\right)$

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	$x.P\left(x\right)$	$x^2.P\left(x\right)$
1	953	0.563	0.56	0.56
2	423	0.250	0.50	1.00
3	194	0.115	0.34	1.03
4	80	0.047	0.19	0.76
5	29	0.017	0.09	0.43
6	9	0.005	0.03	0.19
7	3	0.002	0.01	0.09
8	1	0.001	0.00	0.04
Total	1692	1.000	1.73	4.10

  $\begin{array}{c}\text{Standard}\text{Deviation = }\sqrt{\sum x^{2}p\left(x\right)-{\left[xp\left(x\right)\right]}^2}\\ =\sqrt{4.10-{1.73}^2}\\ =1.05\end{array}$