Chapter 6.1, Problem 47E

Relax! The General Social Survey asked 1676 people how many hours per day they were able to relax. The results are presented in the following table.

Consider these 1676 people to be a population. Let X be the number of hours of relaxation for a person sampled at random from this population.

Construct the probability distribution of X.

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

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

Compute the mean $\mu X.$

Compute the standard deviation $\sigma X.$

To determine

To construct:The probability distribution of the given random variable.

Explanation of Solution

The relaxing time of $1676$ people is summarized as the following table for a general society survey.

  $\begin{array}{cc}\text{Number of Hours}& \text{Frequency}\\ 0& 114\\ 1& 186\\ 2& 336\\ 3& 251\\ 4& 316\\ 5& 231\\ 6& 149\\ 7& 33\\ 8& 60\\ \text{Total}& 1676\end{array}$

Calculation:

The random variable $\left(X\right)$ is considered as the number of hours of relaxation.

To calculate the probability of each value of the random variable, the frequency should be divided by the total number of individuals according to the formula,

  $P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

As an example,

  $P\left(x=0\right)=\frac{114}{1696}=0.0672$

The all calculation can be expressed in a table as follows.

  $\begin{array}{ccc}x& f& P\left(x\right)\\ 0& 114& \frac{114}{1696}=0.0672\\ 1& 186& \frac{186}{1696}=0.1097\\ 2& 336& \frac{336}{1696}=0.1981\\ 3& 251& \frac{251}{1696}=0.1480\\ 4& 316& \frac{316}{1696}=0.1863\\ 5& 231& \frac{231}{1696}=0.1362\\ 6& 149& \frac{149}{1696}=0.0879\\ 7& 33& \frac{33}{1696}=0.0195\\ 8& 60& \frac{60}{1696}=0.0354\end{array}$

The probability distribution can be constructed by the first and third columns of the above table.

  $\begin{array}{cc}x& P\left(x\right)\\ 0& 0.0672\\ 1& 0.1097\\ 2& 0.1981\\ 3& 0.1480\\ 4& 0.1863\\ 5& 0.1362\\ 6& 0.0879\\ 7& 0.0195\\ 8& 0.0354\end{array}$

To determine

To find:The probability to the relaxation time to be more than four hours.

Answer to Problem 47E

The probability to that a person relaxes more than four hours is $0.2789$ .

Explanation of Solution

The probability distribution for relaxing time of $1676$ people is constructed as follows in the part (a).

  $\begin{array}{cccccccccc}x& 0& 1& 2& 3& 4& 5& 6& 7& 8\\ P\left(x\right)& 0.0672& 0.1097& 0.1981& 0.1480& 0.1863& 0.1362& 0.0879& 0.0195& 0.0354\end{array}$

Calculation:

Having relaxation for more than four hours means the random variable $x$ should be greater than $4$ . Hence, the possible values for the case is said to be $5,6,7\text{ and }8$ .

The probability can be expressed in the notation as,

  $P\left(x>4\right)=P\left(5\text{ or }6\text{ or }7\text{ or }8\right)$

Since $x=5,x=6,x=7$ and $x=8$ are mutually exclusive events, the addition rule can be applied.

  $\begin{array}{c}P\left(5\text{ or }6\text{ or }7\text{ or }8\right)=P\left(5\right)+P\left(6\right)+P\left(7\right)+P\left(8\right)\\ =0.1362+0.0879+0.0195+0.0354\\ P\left(x>4\right)=0.2789\end{array}$

Conclusion:

The probability of relaxation hours $\left[P\left(x>4\right)\right]$ to be more than four is found to be $0.2789$ .

To determine

To find:The probability to the relaxation time is zero.

Answer to Problem 47E

The probability to that a person does not take any relax is found to be $0.0672$ .

Explanation of Solution

The probability distribution for relaxing time of $1676$ people is constructed as follows in the part (a).

  $\begin{array}{cccccccccc}x& 0& 1& 2& 3& 4& 5& 6& 7& 8\\ P\left(x\right)& 0.0672& 0.1097& 0.1981& 0.1480& 0.1863& 0.1362& 0.0879& 0.0195& 0.0354\end{array}$Calculation:

When a person does not relax at all, the relaxation time is equal to zero. Hence, the random variable should be $x=0$ .

The relevant probability is calculated in a precious part as,

  $\begin{array}{c}P\left(x=0\right)=\frac{114}{1696}\\ P\left(0\right)=0.0672\end{array}$

Conclusion:

The probability of $x=0$ is found to be $0.0672$ .

To determine

To find: The mean relaxation time.

Answer to Problem 47E

The mean is found to be,

  ${\mu }_X=3.3225$

Explanation of Solution

The probability distribution for relaxing time of $1676$ people is constructed as follows in the part (a).

  $\begin{array}{cccccccccc}x& 0& 1& 2& 3& 4& 5& 6& 7& 8\\ P\left(x\right)& 0.0672& 0.1097& 0.1981& 0.1480& 0.1863& 0.1362& 0.0879& 0.0195& 0.0354\end{array}$Calculation:

The mean of a random variable, or equivalently the expected value is given by the sum of the product of the values and the corresponding probabilities.

  ${\mu }_X=E\left(X\right)=x_1{P}_1+x_2{P}_2+x_3{P}_3+\mathrm{...}$

Here, for each value of $x$ should be multiplied by the probabilities and added.

  $\begin{array}{c}{\mu }_X=\left(0\times 0.0672\right)+\left(1\times 0.1097\right)+\left(2\times 0.1981\right)+\left(3\times \mathrm{0.0.1480}\right)\\ +\left(4\times 0.1863\right)+\left(5\times 0.1362\right)+\left(6\times 0.0879\right)+\left(7\times 0.0195\right)+\left(8\times 0.0354\right)\\ =0+0.1097+0.3962+0.4440+0.7453+0.6810+0.5271+0.1362+0.2890\\ {\mu }_X=3.3225\end{array}$

Conclusion:

The mean number of relaxation time is found to be $3.3225$ .

To determine

To find:The standard deviation of $X$ .

Answer to Problem 47E

The standard deviation is found to be,

  ${\sigma }_X=1.9574$

Explanation of Solution

The probability distribution for relaxing time of $1676$ people is constructed as follows in the part (a).

  $\begin{array}{cccccccccc}x& 0& 1& 2& 3& 4& 5& 6& 7& 8\\ P\left(x\right)& 0.0672& 0.1097& 0.1981& 0.1480& 0.1863& 0.1362& 0.0879& 0.0195& 0.0354\end{array}$Calculation:

The variance of a random variable $\left({\sigma }_X{}^2\right)$ is given by the formula,

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

By constructing a table we can do the calculations clearly using the mean of $3.3225$ .

  $\begin{array}{ccccc}x& x-{\mu }_X& {\left(x-{\mu }_X\right)}^2& P\left(x\right)& {\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)\\ 0& -3.3225& 11.0390& 0.0672& 0.7420\\ 1& -2.3225& 5.3940& 0.1097& 0.5916\\ 2& -1.3225& 1.7490& 0.1981& 0.3465\\ 3& -0.3225& 0.1040& 0.1480& 0.0154\\ 4& 0.6775& 0.4590& 0.1863& 0.0855\\ 5& 1.6775& 2.8140& 0.1362& 0.3833\\ 6& 2.6775& 7.1690& 0.0879& 0.6298\\ 7& 3.6775& 13.5240& 0.0195& 0.2631\\ 8& 4.6775& 21.8790& 0.0354& 0.7740\end{array}$

The sum of right-most column gives the variation of $X$ .

  ${\sigma }_X{}^2=3.8313$

The standard deviation $\left({\sigma }_X\right)$ is the square root of variance. Hence,

  $\begin{array}{c}{\sigma }_X=\sqrt{{\sigma }_X{}^2}\\ =\sqrt{3.8313}\\ {\sigma }_X=1.9574\end{array}$

Conclusion:

The standard deviation is found to be $1.9574$ .