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# Binomial and Normal Variable

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1. A manufacturer of window frames knows from long experience that 5 percent of the production

will have some type of minor defect that will require an adjustment. What is the probability that in a sample of 20 window frames:

b. At least one will need adjustment?

More than two will need adjustment?
2. A study of Furniture Wholesales, Inc. regarding the payment of invoices reveals the time from billing until payment is received follows the normal distribution. The mean time until payment is received is 20 days and the standard deviation is 5 days.

a. What percent of the invoices are paid within 15 days of receipt?

b. What percent of the invoices are paid in more than 28 days?

c. What percent of the invoices are paid in more than 15 days but less than 28 days?

d. The management of Furniture Wholesales wants to encourage their customers to pay their monthly invoices as soon as possible. Therefore, it announced that a 2 percent reduction in price would be in effect for customers who pay within 7 working days of the receipt of the invoice. What percent of customers will earn this discount?

3. The annual commissions earned by sales representatives of Machine Products Inc., a manufacturer of light machinery, follow the normal distribution. The mean yearly amount earned is \$40,000 and the standard deviation is \$5,000.

a. What percent of the sales representatives earn more than \$42,000 per year?

b. What percent of the sales representatives earn between \$32,000 and \$42,000?

c. What percent of the sales representatives earn between \$32,000 and \$35,000?

d. The sales manager wants to award the sales representatives who earn the largest commissions a bonus of \$1,000. He can award a bonus to 20 percent of the representatives. What is the cutoff point between those who earn a bonus and those who do not?

https://brainmass.com/statistics/probability/binomial-and-normal-variable-99784

#### Solution Preview

Probability questions need full explanations

1. A manufacturer of window frames knows from long experience that 5 percent of the production will have some type of minor defect that will require an adjustment. What is the probability that in a sample of 20 window frames:

b. At least one will need adjustment?

c. More than two will need adjustment?

Let E denotes the event that a unit requires minor adjustment

Given P(E) = 0.05
Then P(Ec) = 0.95 is the probability of the event that the unit requires no adjustment.

Let X denote the random variable that number of units that requires adjustment.
Here X can take values X= (0,1,2,.............20)

Here we need P(X= 0)
The probability that no unit need adjustment can be represented as
P(Ec ∩ Ec ∩ Ec ∩ Ec ∩ Ec ∩ ...... Ec) = 0.9520 =0.358486

b. At least one will need adjustment?

P(At least one will need adjustment ) =P(X ≥ 1)
= 1 - P (None will need adjustment)
= 1 - 0.9520 = 0.641514

c. More than two will need adjustment?

P(More than two will need adjustment ) = P(X≥ 2 )
= 1-P(X<2)
= 1- P(X=0)-P(X=1)
P(X = 0) = 0.9520
= ...

#### Solution Summary

The solution the details for probability calculation of binomial and normal variable.

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