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Chapter 6: Problem 40

Tommy Wait, a minor league baseball pitcher, is notorious for taking an excessive amount of time between pitches. In fact, his times between pitches are normally distributed with a mean of 36 seconds and a standard deviation of \(2.5\) seconds. What percentage of his times between pitches are a. longer than 39 seconds? b. between 29 and 34 seconds?

Short Answer

Expert verified
a. The percentage of times Tommy Wait takes longer than 39 seconds between pitches. b. The percentage of times Tommy Wait takes between 29 and 34 seconds between pitches.

Step by step solution

01

Standardize the Given Time Value for Question A

For any normal distribution, any value X can be standardized using the formula z = (X - 渭)/蟽, where 渭 is the mean and 蟽 is the standard deviation. For the time value of 39 seconds, this would be z = (39 - 36)/2.5, calculate this to obtain the z-value.
02

Find the Corresponding Probability for the Calculated Z-Value

Use a standard normal distribution table (also known as a z-table) to find the probability corresponding to the calculated z-value from Step 1. Since the question asks for the proportion of times longer than 39 seconds, look for the right tail area in the z-table.
03

Standardize the Given Time Values for Question B

Repeat the process from Steps 1-2 to handle the second question, but this time with two time values (29 and 34 seconds). The difference is that this time, we are looking for the proportion of times in between these two values. So after standardizing both these times, find the corresponding probabilities from the z-table and subtract the smaller probability from the larger one to obtain the proportion of times between these values.
04

Convert Probabilities to Percentages

The probabilities obtained from the table might be between 0 and 1. To express these as percentages, multiply the probabilities by 100.

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Most popular questions from this chapter

For a binomial probability distribution, \(n=25\) and \(p=.40\). a. Find the probability \(P(8 \leq x \leq 13)\) by using the table of binomial probabilities (Table I of Appendix C). b. Find the probability \(P(8 \leq x \leq 13)\) by using the normal distribution as an approximation to the binomial distribution. What is the difference between this approximation and the exact probability calculated in part a?

Find the area under the standard normal curve a. between \(z=0\) and \(z=1.95\) b. between \(z=0\) and \(z=-2.05\) c. between \(z=1.15\) and \(z=2.37\) d. from \(z=-1.53\) to \(z=-2.88\) e. from \(z=-1.67\) to \(z=2.24\)

Fast Auto Service guarantees that the maximum waiting time for its customers is 20 minutes for oil and lube service on their cars. It also guarantees that any customer who has to wait longer than 20 minutes for this service will receive a \(50 \%\) discount on the charges. It is estimated that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is \(2.4\) minutes. Suppose the time taken for oil and lube service on a car follows a normal distribution. a. What percentage of the customers will receive the \(50 \%\) discount on their charges? b. Is it possible that a car may take longer than 25 minutes for oil and lube service? Explain

What are the parameters of the normal distribution?

Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is \(2.4\) minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a \(50 \%\) discount on the charges. The company wants to limit this discount to at most \(5 \%\) of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.
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