Concept explainers
a.
To find the mean of the given probability distribution.
a.
Answer to Problem 2RE
Mean = 8.37
Explanation of Solution
Given:
x | P ( x ) |
6 | 0.21 |
7 | 0.12 |
8 | 0.29 |
9 | 0.11 |
10 | 0.01 |
11 | 0.26 |
Concept used:
x | P ( x ) | |
6 | 0.21 | 1.26 |
7 | 0.12 | 0.84 |
8 | 0.29 | 2.32 |
9 | 0.11 | 0.99 |
10 | 0.01 | 0.1 |
11 | 0.26 | 2.86 |
Sum | 1 | 8.37 |
Since the
Therefore, the mean of the given distribution is 8.37.
b.
To find the Variance of the given probability distribution.
b.
Answer to Problem 2RE
Variance = 3.3131
Explanation of Solution
Given:
x | P ( x ) |
6 | 0.21 |
7 | 0.12 |
8 | 0.29 |
9 | 0.11 |
10 | 0.01 |
11 | 0.26 |
Concept used:
x | P ( x ) | ||
6 | 0.21 | 1.26 | 7.56 |
7 | 0.12 | 0.84 | 5.88 |
8 | 0.29 | 2.32 | 18.56 |
9 | 0.11 | 0.99 | 8.91 |
10 | 0.01 | 0.1 | 1 |
11 | 0.26 | 2.86 | 31.46 |
Sum | 1 | 8.37 | 73.37 |
From the above table:
Therefore, the variance is,
c.
To find the Standard Deviation of the given probability distribution.
c.
Answer to Problem 2RE
Standard Deviation = 1.82
Explanation of Solution
Given:
x | P ( x ) |
6 | 0.21 |
7 | 0.12 |
8 | 0.29 |
9 | 0.11 |
10 | 0.01 |
11 | 0.26 |
Concept used:
x | P ( x ) | ||
6 | 0.21 | 1.26 | 7.56 |
7 | 0.12 | 0.84 | 5.88 |
8 | 0.29 | 2.32 | 18.56 |
9 | 0.11 | 0.99 | 8.91 |
10 | 0.01 | 0.1 | 1 |
11 | 0.26 | 2.86 | 31.46 |
Sum | 1 | 8.37 | 73.37 |
From the above table:
Therefore, the variance is,
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- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillCalculus For The Life SciencesCalculusISBN:9780321964038Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.Publisher:Pearson Addison Wesley,