Question

Apollo Gates Inc, one of the largest makers of sliding gate openers in the country based out of Dallas uses rolling wheels for their gates (model AB) that are targeted to be 5 inches in diameter. The acceptable limits of rolling wheels are as follows: on the higher end 5.02 inches and the lower end 4.98 inches. Any wheels that do not meet these specifications are known to cause catastrophic failure and lead to significant damages for which company could be held liable. The quality assurance testing for these wheels has shown that the manufacturing of the wheels follows a normal distribution with a mean of 5.0045 inches and a standard deviation of 0.018 inches. What is the probability that the wheels …. (show your calculations and write-up below each question part, there are 5 parts in this question) 5 points

Please provide EXCEL steps

1) Will be between the target of 5 inches and the actual value

2) Will be between the target and the upper limit

3) Above the upper limit

4) Below lower Limit

5) Of all wheels, 90% of them are greater than what value?

Answer 1

Answer:

To answer these questions we first have to normalize the distribution to be able to use a normal table. This means that for any given radius X, we have to calculate the standard normal value:

Z = (X - m)/s

Where m is the mean and s is the standard deviation of the distribution.

1)

The first question basically asks what is the probability that a given wheel with radius X will be such that 5 < X < 5.0045.

So we have to find P( 5 < X < 5.0045 ) = P( -0.25 < (X - 5.0045)/0.018 < 0 ) = P( -0.25 < Z < 0 ) = 0.0987.

So the probability that the wheels are between the target and the actual value of the mean is of 9.87%

2)

This next question asks to find P( 5 < X < 5.02 ). Again we standardize as follows:

P( -0.25 < (X - 5.0045)/0.018 < 0.0861 ) = P( -0.25 < Z < 0.0861 ) = P( -0.25 < Z) + P( Z < 0.0861 ) = 0.0987 + 0.0359 = 0.1346.

3)

Now we have to find P( 5.02 < X ). Which is found as follows:

P( 5.02 < X ) = P( 0.0861 < Z ) = 0.5 - 0.0359 = 0.4641

4)

This is:

P( X < 4.98 ) = P ( Z < -1.3611 ) = 0.0869

5)

Now we are asked to find x such that P( x < X ) = 0.9.

Using a normal table we find that P( -1.003 < Z ) = 0.8997 which is the closest we can get to 0.9 on a normal table.

This means that of all the values that Z can take, around 90% of them are above -1.003.

This implies that of all the values X can take, then around 90% of them are above (-1.003)0.018 + 5.0045 = 4.9864.