You are the manager of a small company that produces industrial parts

Mary Keefe

Mary Keefe

Answered question

2021-12-07

You are the manager of a small company that produces industrial parts for automobile assembly.
One of your products is too technical and therefore takes longer than the other parts to produce. As a manager, you want to make sure that the quantity you produced is more or less in line with demand.
Based on historical data, he got the following:
X300400500600Probability0.200.300.350.15
a) Explain in words what the random variable is in this case.
b) What is the expected value of the random variable?
c) What is the variance of the random variable?

Answer & Explanation

scomparve5j

scomparve5j

Beginner2021-12-08Added 38 answers

Step 1
The formula to calculate the expected value and the variance of the random variable is:
E(X)=xP(x)
V(X)=x2P(x)[xP(x)]2
=x2P(x)[E(X)]2
Step 2
a) As the manager wants to check whether the quantity produced is meeting the demand, the random variable (X) in this case is, "X: the quantity of product produced that is in line with the demand".
Step 3
b) The expected value of the random variable can be calculated as follows:
xP(x)xP(x)3000.20604000.301205000.351756000.1590SUM1.00445.00
E(X)=xP(x)
=445
Therefore, the expected value of X is 445.
Step 4
c) The variance of the random variable can be calculated as follows:
xP(x)x2P(x)3000.20180004000.30480005000.35875006000.1554000SUM1.00207500
V(X)=x2P(x)[E(X)]2
=207500[445]2
=9475
Therefore, the variance of X is 9475.

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