Chapter 5.1, Problem 49E

a.

To determine

Construct the probability distribution for the random variable X.

Answer to Problem 49E

The probability distribution for the random variable X is,

x	P(x)
1	0.1278
2	0.1241
3	0.1236
4	0.1222
5	0.1227
6	0.1247
7	0.1266
8	0.1283

Explanation of Solution

Calculation:

The table represents the numbers of students enrolled in grades 1 through 8 in public schools in United States. The random variable X denotes the grade of a student sampled at random from the population.

Here, the total number of students is 29,343. The corresponding probabilities of the random variable X are obtained by dividing the frequency of students in each grade (f) by total frequency (N).

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

For x=1:

$\begin{array}{c}P\left(1\right)=\frac{3,750}{29,343}\\ =0.1278\end{array}$

Similarly the remaining probabilities are obtained as follows:

x	$\frac{f}{N}$	P(x)
1	$\frac{3,750}{29,343}$	0.1278
2	$\frac{3,640}{29,343}$	0.1241
3	$\frac{3,627}{29,343}$	0.1236
4	$\frac{3,585}{29,343}$	0.1222
5	$\frac{3,601}{29,343}$	0.1227
6	$\frac{3,660}{29,343}$	0.1247
7	$\frac{3,715}{29,343}$	0.1266
8	$\frac{3,765}{29,343}$	0.1283
Total		1.0000

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

b.

To determine

Find the probability that the student is in fourth grade.

Answer to Problem 49E

The probability that the student is in fourth grade is 0.1222.

Explanation of Solution

The table obtained in part (a) represents the probability distribution of the random variable X, the number of students enrolled in each grade.

The probability that the student is in fourth grade is the probability at $x=4$.

From the probability distribution table, the probability at $x=4$ is 0.1222.

Thus, the probability that the student is in fourth grade is 0.1222.

c.

To determine

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

Answer to Problem 49E

The probability that the student is in seventh or eighth grade is 0.2549.

Explanation of Solution

Calculation:

The probability that the student is in seventh or eighth grade is the sum of the probabilities at the points $x=7\text{ and }x=8$.

$P\left(\text{Student is in 7 or 8}\right)=P\left(x=7\right)+P\left(x=8\right)$

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

$\begin{array}{c}P\left(\text{Student is in 7 or 8}\right)=P\left(x=7\right)+P\left(x=8\right)\\ =0.1266+0.1283\\ =0.2549\end{array}$

Thus, the probability that the student is in seventh or eighth grade is 0.2549.

d.

To determine

Find the mean.

Answer to Problem 49E

The mean value is 4.51.

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)	$x\cdot P\left(x\right)$
1	0.1278	0.13
2	0.1241	0.25
3	0.1236	0.37
4	0.1222	0.49
5	0.1227	0.61
6	0.1247	0.75
7	0.1266	0.89
8	0.1283	1.03
Total	1.0000	4.51

Thus, the mean value is 4.51.

f.

To determine

Find the standard deviation.

Answer to Problem 49E

The standard deviation is 2.31.

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)	$\left(x-{\mu }_X\right)$	${\left(x-{\mu }_X\right)}^2$	${\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)$
1	0.1278	–3.51	12.3201	1.5745
2	0.1241	–2.51	6.3001	0.7815
3	0.1236	–1.51	2.2801	0.2818
4	0.1222	–0.51	0.2601	0.0318
5	0.1227	0.49	0.2401	0.0295
6	0.1247	1.49	2.2201	0.2769
7	0.1266	2.49	6.2001	0.7850
8	0.1283	3.49	12.1801	1.5628
Total	1.0000	–0.08	42.0008	5.3238

Therefore,

${\sigma }_X^2=5.3238$

Thus, the variance is 5.3238.

The standard deviation is,

$\begin{array}{c}{\sigma }_X=\sqrt{5.3238}\\ =2.31\end{array}$

That is, the standard deviation is 2.31.