Chapter 5.1, Problem 47E

a.

To determine

Construct the probability distribution for the random variable X.

Answer to Problem 47E

The probability distribution for the random variable X is,

x	P(x)
0	0.0680
1	0.1110
2	0.2005
3	0.1498
4	0.1885
5	0.1378
6	0.0889
7	0.0197
8	0.0358

Explanation of Solution

Calculation:

The table represents the results of the survey in which 1,676 people are asked how many hours per day they were able to relax. The random variable X denotes the number of hours of relaxation for a person sampled at random from the population.

Here, the total number of people is 1,676. 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=0:

$\begin{array}{c}P\left(0\right)=\frac{114}{1,676}\\ =0.0680\end{array}$

Similarly the remaining probabilities are obtained as follows:

x	$\frac{f}{N}$	P(x)
0	$\frac{114}{1,676}$	0.0680
1	$\frac{186}{1,676}$	0.1110
2	$\frac{336}{1,676}$	0.2005
3	$\frac{251}{1,676}$	0.1498
4	$\frac{316}{1,676}$	0.1885
5	$\frac{231}{1,676}$	0.1378
6	$\frac{149}{1,676}$	0.0889
7	$\frac{33}{1,676}$	0.0197
8	$\frac{60}{1,676}$	0.0358
Total		1.0000

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

b.

To determine

Find the probability that a person relaxes more than 4 hours per day.

Answer to Problem 47E

The probability that a person relaxes more than 4 hours per day is 0.282.

Explanation of Solution

Calculation:

The table represents the probability distribution of the random variable X, the number of hours of relaxation for a person.

The probability that a person relaxes more than 4 hours per day is the sum of the probabilities at the points $x=5,x=6,x=7\text{ and }x=8$.

$P\left(x>4\right)=P\left(x=5\right)+P\left(x=6\right)+P\left(x=7\right)+P\left(x=8\right)$

Substituting the values from the probability distribution table obtained in part (a),

$\begin{array}{c}P\left(x>4\right)=P\left(x=5\right)+P\left(x=6\right)+P\left(x=7\right)+P\left(x=8\right)\\ =0.1378+0.0889+0.0197+0.0358\\ \approx 0.282\end{array}$

Thus, the probability that a person relaxes more than 4 hours per day is 0.282.

c.

To determine

Find the probability that the person doesn’t relax at all.

Answer to Problem 47E

The probability that the person doesn’t relax at all is 0.068.

Explanation of Solution

Calculation:

The probability that the person doesn’t relax at all is the probability at $x=0$.

From the probability distribution table, the probability at $x=0$ is 0.068.

Thus, the probability that the person doesn’t relax at all is 0.068.

d.

To determine

Find the mean.

Answer to Problem 47E

The mean value is 3.36.

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(x)	$x\cdot P\left(x\right)$
0	0.0680	0.00
1	0.1110	0.11
2	0.2005	0.40
3	0.1498	0.45
4	0.1885	0.75
5	0.1378	0.69
6	0.0889	0.53
7	0.0197	0.14
8	0.0358	0.29
Total	1.0000	3.36

Thus, the mean value is 3.36.

f.

To determine

Find the standard deviation.

Answer to Problem 47E

The standard deviation is 1.97.

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(x)	$\left(x-{\mu }_X\right)$	${\left(x-{\mu }_X\right)}^2$	${\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)$
0	0.0680	–3.3621	11.3037	0.7687
1	0.1110	–2.3621	5.5795	0.6193
2	0.2005	–1.3621	1.8553	0.3720
3	0.1498	–0.3621	0.1311	0.0196
4	0.1885	0.6379	0.4069	0.0767
5	0.1378	1.6379	2.6827	0.3697
6	0.0889	2.6379	6.9585	0.6186
7	0.0197	3.6379	13.2343	0.2607
8	0.0358	4.6379	21.5101	0.7701
Total	1.0000	5.7411	63.6622	3.8754

Therefore,

${\sigma }^2=3.8754$

Thus, the variance is 3.8754.

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{3.8754}\\ =1.97\end{array}$

That is, the standard deviation is 1.97.