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Stels [109]
3 years ago
10

Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a service facility. Suppose they ar

e normal random variables with means of 50 minutes, 60 minutes, and 40 minutes, respectively. The standard deviations are 15 minutes, 20 minutes, and 10 minutes, respectively. a Suppose X1, X2, and X3 are independent. All three repairs must be completed on a given object. What is the mean and variance of the total repair time for this object.
Mathematics
1 answer:
belka [17]3 years ago
8 0

Answer:

The mean of the total repair time is 150 minutes.

The variance of the total repair time is 725 minutes^2.

Step-by-step explanation:

To solve this problem, we have to use the properties of the mean and the variance. Our random variable is the sum of 3 normal variables.

In the case, for the mean, we have that the mean of the sum of 3 normal variables is equal to the sum of the mean of the 3 variables:

y=x_1+x_2+x_3 \\\\E(y)=E(x_1+x_2+x_3)=E(x_1)+E(x_2)+E(x_3)\\\\E(y)=50+60+40=150

For the variance, we apply the property for the sum of independent variables (the correlation between the variables is 0):

V(y)=V(x_1)+V(x_2)+V(x_3)\\\\V(y)=s_1^2+s_2^2+s_3^2\\\\V(y)=15^2+20^2+10^2\\\\V(y)=225+400+100\\\\V(y)=725

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Agata [3.3K]

1. expand using place value.

337,060 = 300,000 + 30,000 + 7,000 + 000 + 60 + 0


Next, we will use exponents:

300,000 = 3 * 10^5

30,000 = 3 * 10^4

7,000 = 7 * 10^3

000 = 0 * 10^2

60 = 6 * 10

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then after combing these exponents, we can write the number as:

337,060 = 3 * 10^5 + 3 * 10^4 + 7 * 10^3 + 0 * 10^2 + 6 * 10 + 0 * 10^0


Finally, removing the meaningless zeroes, we would end up with:

337,060 = 3 * 10^5 + 3 * 10^4 + 7 * 10^3 + 6 * 10


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