Chapter 6.1, Problem 49E

School days: The following table presents the numbers of students enrolled in grades 1 through 8 in public schools in the United States.

Consider these students to be a population. Let X be the grade of a student randomly chosen from this population.

Construct the probability distribution of X.

Find the probability that the student is in fourth grade.

Find the probability that the student is seventh or eighth grade.

Compute the mean $\mu X.$

Compute the standard $\sigma X.$

To determine

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

Explanation of Solution

The enrollment of students from grade $1$ to grade $8$ is given in the following table. The frequency is in thousands. .

  $\begin{array}{cc}\text{Grade}& \text{Frequency}\\ 1& 3750\\ 2& 3640\\ 3& 3627\\ 4& 3585\\ 5& 3601\\ 6& 3660\\ 7& 3715\\ 8& 3765\\ \text{Total}& 29,343\end{array}$

Calculation:

The random variable $\left(X\right)$ is considered as the grade.

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,

  $P\left(x=1\right)=\frac{3,750,000}{29,343,000}=0.1278$

The all calculation can be expressed in a table as follows. Because both values in the numerator and the denominator is in thousands, in the division those are cancelled out.

  $\begin{array}{ccc}x& f& P\left(x\right)\\ 1& 3750& \frac{3750}{29,343}=0.1278\\ 2& 3640& \frac{3640}{29,343}=0.1241\\ 3& 3627& \frac{3627}{29,343}=0.1236\\ 4& 3585& \frac{3585}{29,343}=0.1222\\ 5& 3601& \frac{3601}{29,343}=0.1227\\ 6& 3660& \frac{3660}{29,343}=0.1247\\ 7& 3715& \frac{3715}{29,343}=0.1266\\ 8& 3765& \frac{3765}{29,343}=0.1283\end{array}$

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

  $\begin{array}{cc}x& P\left(x\right)\\ 1& 0.1278\\ 2& 0.1241\\ 3& 0.1236\\ 4& 0.1222\\ 5& 0.1227\\ 6& 0.1247\\ 7& 0.1266\\ 8& 0.1283\end{array}$

To determine

To find: The probability to a selected student is from grade four.

Answer to Problem 49E

The probability that the student is in fourth grade is found to be $0.1222$ .

Explanation of Solution

The probability distribution for the grade of $29,343$ thousand students is constructed as follows in the part (a).

  $\begin{array}{ccccccccc}x& 1& 2& 3& 4& 5& 6& 7& 8\\ P\left(x\right)& 0.1278& 0.1241& 0.1236& 0.1222& 0.1227& 0.1247& 0.1266& 0.1283\end{array}$

Calculation:

When a student is fin grade four, the random variable $x$ is equal to four; $x=0$ .

The relevant probability is calculated in a precious part as,

  $\begin{array}{c}P\left(x=4\right)=\frac{3,585,000}{29,343,000}\\ P\left(4\right)=0.1222\end{array}$

Conclusion:

The probability of $x=4$ is found to be $0.1222$ .

To determine

To find: The probability to a selected student is in grade seven or eight.

Answer to Problem 49E

The probability that the student is seventh or eighth grade is found to be $0.2549$ .

Explanation of Solution

The probability distribution for the grade of $29,343$ thousand students is constructed as follows in the part (a).

  $\begin{array}{ccccccccc}x& 1& 2& 3& 4& 5& 6& 7& 8\\ P\left(x\right)& 0.1278& 0.1241& 0.1236& 0.1222& 0.1227& 0.1247& 0.1266& 0.1283\end{array}$

Calculation:

Same student cannot enroll to two different grades. Hence, being a grade seven student and being a grade eight student are two mutually exclusive events.

Therefore, the probability for this combination can be written as,

  $P\left(7\text{ or }8\right)$

By the addition rule, this probability should be equal to

  $P\left(x=7\right)+P\left(x=8\right)$

The total probability can be determined as,

  $\begin{array}{c}P\left(7\right)+P\left(8\right)=0.1266+0.1283\\ =0.2549\end{array}$

Conclusion:

The probability of $P\left(7\text{ or }8\right)$ is found to be $0.2549$ .

To determine

To find: The mean of grade of the student.

Answer to Problem 49E

The mean is found to be,

  ${\mu }_X=4.5101$

Explanation of Solution

The probability distribution for the grade of $29,343$ thousand students is constructed as follows in the part (a).

  $\begin{array}{ccccccccc}x& 1& 2& 3& 4& 5& 6& 7& 8\\ P\left(x\right)& 0.1278& 0.1241& 0.1236& 0.1222& 0.1227& 0.1247& 0.1266& 0.1283\end{array}$

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.

  $\begin{array}{c}{\mu }_X=\left(1\times 0.1278\right)+\left(2\times 0.1241\right)+\left(3\times 0.1236\right)\\ +\left(4\times 0.1222\right)+\left(5\times 0.1227\right)+\left(6\times 0.1247\right)+\left(7\times 0.1266\right)+\left(8\times 0.1283\right)\\ =0.1278+0.2481+0.3708+0.4887+0.6136+0.7484+0.8862+1.0265\\ {\mu }_X=4.5101\end{array}$

Conclusion:

The mean is found to be $4.5101$ .

To determine

To find: The standard deviation of $X$ .

Answer to Problem 49E

The standard deviation is found to be,

  ${\sigma }_X=2.3073$

Explanation of Solution

The probability distribution for the grade of $29,343$ thousand students is constructed as follows in the part (a).

  $\begin{array}{ccccccccc}x& 1& 2& 3& 4& 5& 6& 7& 8\\ P\left(x\right)& 0.1278& 0.1241& 0.1236& 0.1222& 0.1227& 0.1247& 0.1266& 0.1283\end{array}$

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 $4.5101$ .

  $\begin{array}{ccccc}x& x-{\mu }_X& {\left(x-{\mu }_X\right)}^2& P\left(x\right)& {\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)\\ 1& -3.5101& 12.3211& 0.1278& 1.5746\\ 2& -2.5101& 6.3008& 0.1241& 0.7816\\ 3& -1.5101& 2.2805& 0.1236& 0.2819\\ 4& -0.5101& 0.2602& 0.1222& 0.0318\\ 5& 0.4899& 0.2399& 0.1227& 0.0294\\ 6& 1.4899& 2.2197& 0.1247& 0.2769\\ 7& 2.4899& 6.1994& 0.1266& 0.7849\\ 8& 3.4899& 12.1791& 0.1283& 1.5627\end{array}$

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

  ${\sigma }_X{}^2=5.3238$

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

  $\begin{array}{c}{\sigma }_X=\sqrt{{\sigma }_X{}^2}\\ =\sqrt{5.3238}\\ {\sigma }_X=2.3073\end{array}$

Conclusion:

The standard deviation is found to be $2.3073$ .