Concept explainers
a.
Construct the
a.

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,
For x=1:
Similarly the remaining probabilities are obtained as follows:
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 | |
Total | 1.0000 |
Thus, the discrete probability distribution for the random variable X is obtained.
b.
Find the probability that the student is in fourth grade.
b.

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
From the probability distribution table, the probability at
Thus, the probability that the student is in fourth grade is 0.1222.
c.
Find the probability that the student is in seventh or eighth grade.
c.

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
Substituting the values from the probability distribution table obtained in part (a),
Thus, the probability that the student is in seventh or eighth grade is 0.2549.
d.
Find the
d.

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,
The mean of the random variable is obtained as given below:
x | P(x) | |
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.
Find the standard deviation.
f.

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,
Where
The variance of the random variable X is obtained using the following table:
x | P(x) | |||
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,
Thus, the variance is 5.3238.
The standard deviation is,
That is, the standard deviation is 2.31.
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Chapter 5 Solutions
ALEKS 360 ESSENT. STAT ACCESS CARD
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