Chapter 5, Problem 4RE

a.

To determine

Construct the probability distribution for the random variable X.

Answer to Problem 4RE

The probability distribution for the random variable X is,

x	P(x)
1	0.5632
2	0.2500
3	0.1147
4	0.0473
5	0.0171
6	0.0053
7	0.0018
8	0.0006

Explanation of Solution

Calculation:

The given table represents the frequency distribution of the number of AP tests taken by students who took one or more AP tests. The random variable X denotes the number of exams taken by a student who took one or more AP tests.

Here, the total number of people is 1,692. The corresponding probabilities of the random variable X are obtained by dividing the frequency of each hour (f) by total frequency (N).

That is, $P\left(x\right)=\frac{f}{N}$

For x=1:

$\begin{array}{c}P\left(1\right)=\frac{953}{1,692}\\ =0.5632\end{array}$

Similarly the remaining probabilities are obtained as follows:

x	$\frac{f}{N}$	P(x)
1	$\frac{953}{1,692}$	0.5632
2	$\frac{423}{1,692}$	0.2500
3	$\frac{194}{1,692}$	0.1147
4	$\frac{80}{1,692}$	0.0473
5	$\frac{29}{1,692}$	0.0171
6	$\frac{9}{1,692}$	0.0053
7	$\frac{3}{1,692}$	0.0018
8	$\frac{1}{1,692}$	0.0006
Total		1.0000

Thus, the discrete probability distribution for the random variable X is obtained.

b.

To determine

Find the probability that a student took exactly one exam.

Answer to Problem 4RE

The probability that a student took exactly one exam is 0.5632.

Explanation of Solution

Calculation:

The table represents the probability distribution of the random variable X, the exams taken by a student who took one or more AP tests.

The probability that a student took exactly one exam is the probability at the point $x=1$. That is,

$P\left(x=1\right)=0.5632$

Thus, the probability that a student took exactly one exam is 0.5632.

c.

To determine

Find the mean.

Answer to Problem 4RE

The mean value is 1.7312.

Explanation of Solution

Calculation:

The formula for the mean of a discrete random variable is,

$\begin{array}{c}E\left(X\right)={\mu }_X\\ ={\displaystyle \sum x\cdot P\left(x\right)}\end{array}$

The mean of the random variable is obtained as given below:

x	$P\left(x\right)$	$x\cdot P\left(x\right)$
1	0.5632	0.5632
2	0.2500	0.5000
3	0.1147	0.3441
4	0.0473	0.1892
5	0.0171	0.0855
6	0.0053	0.0318
7	0.0018	0.0126
8	0.0006	0.0048
Total	1.0000	1.7312

Thus, the mean value is 1.7312.

d.

To determine

Find the standard deviation.

Answer to Problem 4RE

The standard deviation is 1.049.

Explanation of Solution

Calculation:

The standard deviation of the random variable X is obtained by taking the square root of variance.

The formula for the variance of the discrete random variable X is,

${\sigma }_X^2={\displaystyle \sum \left[{\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)\right]}$

Where ${\mu }_X={\displaystyle \sum \left[x\cdot P\left(x\right)\right]}$ represents the mean of the random variable X.

The variance of the random variable X is obtained using the following table:

x	$P\left(x\right)$	$\left(x-{\mu }_X\right)$	${\left(x-{\mu }_X\right)}^2$	${\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)$
1	0.5632	–0.7312	0.535	0.3011
2	0.2500	0.2688	0.072	0.0181
3	0.1147	1.2688	1.610	0.1847
4	0.0473	2.2688	5.147	0.2435
5	0.0171	3.2688	10.685	0.1827
6	0.0053	4.2688	18.223	0.0966
7	0.0018	5.2688	27.760	0.0500
8	0.0006	6.2688	39.298	0.0236
Total	1.0000	22.1504	103.330	1.1001

Therefore,

${\sigma }^2=1.1001$

Thus, the variance is 1.1001.

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{1.1001}\\ =1.049\end{array}$

That is, the standard deviation is 1.049.