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Statistics problems dealing with probability

1. All Seasons Plumbing has two service trucks which frequently break down. If the probability the first truck is available is .75, the probability the second truck is available is .50, and the probability that both trucks are available is .30, what is the probability neither truck is available?

2. Textbook authors and publishers work very hard to minimize the number of errors in a text. However, some errors are unavoidable. Mr. J. A. Carmen, statistics editor, reports that the mean number of errors per chapter is 0.8. What is the probability that there are less than 2 errors in a particular chapter?

3. A recent issue of Bride Magazine suggested that couples planning their wedding should expect two-thirds of those who are sent an invitation to respond that they will attend. Rich and Stacy are planning to be married later this year. They plan to send 197 invitations.

a) How many guests would you expect to accept the invitation?
b) What is the standard deviation?
c) What is the probability 140 or more will accept the invitation?
d) What is the probability exactly 140 will accept the invitation?

4. A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the weight of the boxes, a few were checked. The mean weight was 20.4 pounds, the standard deviation 0.5 pounds. How many boxes must the processor sample to be 95 percent confident that the sample mean does not differ from the population mean by more than 0.2 pounds?

Solution Preview

1. Denote by A the event that the first truck was available, and denote B by the event that the second truck is available. We know that P(A) = 0.75 and P(B) = 0.5, P(AB) = 0.3.

So, P(P U B) = P(A) + P(B) - P(AB) = 0.75 + 0.5 - 0.3 = 0.95

So, P(barA bar B) = 1 - P(A U B) = 1 - 0.95 = 0.05

So, the probability neither truck is available is 0.05.

2. We assume that the number of errors X in a text follows Poisson distribution, namely:

P(X=k) = (1/k!)*e^-λ*(λ^k),k = 0,1,2,3,...

By given ...

Solution Summary

This solution gives a clear, detailed explanation of how to accurately solve four statistics problems, all of which concern probability. The expert provides both calculations and equations, as well as a text explanation of how to achieve the answer.