Question

A highway engineer knows that his crew can lay 5 miles of highway on a clear day, 2 miles on a rainy day, and only 1 mile on a snowy day. Suppose the probabilities are as follows: A clear day: .6, a rainy day: .3, a snowy day: .1. What are the mean and variance

Answer 1

Answer:

Mean = 3.7

Variance = 2.61

Step-by-step explanation:

From the data given; we can represent our table into table format for easier solution and better understanding.

Given that:

A highway engineer knows that his crew can lay 5 miles of highway on a clear day, 2 miles on a rainy day, and only 1 mile on a snowy day

Let X represent the crew;

P(X) represent their respective probabilities

                   clear  day           rainy day            snowy day

X                  5                          2                           1

P(X)              0.6                       0.3                       0.1

From Above; we can determine our X*P(X) and X²P(X)

Let have the two additional columns to table ; we have

X                      P(X)                      X*P(X)                       X²P(X)

5                        0.6                          3                               15

2                        0.3                        0.6                             1.2

1                         0.1                          0.1                            0.1

Total                  1.0                        3.7                             16.3

The mean $\mu$ can be calculated by using the formula:

$\sum \limits ^n _{i=1}X_i P(X_i)$

Therefore ; mean $\mu$ = 3.7

Variance $\sigma^2 = \sum \limits ^n _{i=1}X^2_i P(X_i)- \mu^2$

Variance = 16.3 -3.7²

Variance = 16.3 - 13.69

Variance = 2.61