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Chapter 6: Problem 26

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) Approximately 61.51% of Tommy's pitches take more than 39 seconds. b) Roughly 20.93% of his pitches take between 29 and 34 seconds.

Step by step solution

01

Calculating Z-scores

Z-score is a measure that describes a value's relationship to the mean of a group of values. It's used to compare the two different measurements that may have distinct means or standard deviations. You can calculate the Z-score by using formula: \[Z = \frac{(X - \mu)}{\sigma}\] where \(X\) is the data element, \(\mu\) is the population mean and \(\sigma\) is the standard deviation. For part 'a' the Z score is \(\frac{(39-36)}{2.5} = 1.2\) and for part 'b' Z scores are \(\frac{(29-36)}{2.5} = -2.8\) and \(\frac{(34-36)}{2.5} = -0.8\)
02

Finding the Probability for 'a'

Reflect the Z score into the Standard Normal Distribution Table which shows that a Z score of 1.2 corresponds to 0.3849 or 38.49% (since areas under curves reflect probability directly). However, since the question asks the time that is more than 39 seconds, we look for the probability above 1.2 in the Z distribution which is \(1- 0.3849 = 0.6151\) or 61.51%.
03

Finding the Probability for 'b'

Unlike in step 2, we aren't looking for the probability beyond a certain point, but between two times; 29 and 34 seconds. You will need to find the difference of the probabilities in the Standard Normal Distribution Table of 'b's two Z scores. The probability for Z=-2.8 is 0.0026 and for Z=-0.8 is 0.2119. The difference is \(0.2119 - 0.0026 = 0.2093\) or 20.93%.

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Most popular questions from this chapter

A variation of a roulette wheel has slots that are not of equal size. Instead, the width of any slot is proportional to the probability that a standard normal random variable \(z\) takes on a value between \(a\) and \((a+.1)\), where \(a=-3.0,-2.9,-2.8, \ldots, 2.9,3.0 .\) In other words, there are slots for the intervals \((-3.0,-2.9),(-2.9,-2.8)\), \((-2.8,-2.7)\) through \((2.9,3.0)\). There is one more slot that represents the probability that \(z\) falls outside the interval \((-3.0,3.0)\). Find the following probabilities. a. The ball lands in the slot representing \((.3, .4)\). b. The ball lands in any of the slots representing \((-.1, .4)\). c. In at least one out of five games, the ball lands in the slot representing \((-.1, .4)\). d. In at least 100 out of 500 games, the ball lands in the slot representing \((.4, .5)\)

What are the parameters of the normal distribution?

Find the following areas under a normal distribution curve with \(\mu=20\) and \(\sigma=4\) a. Area between \(x=20\) and \(x=27\) b. Area from \(x=23\) to \(x=26\) c. Area between \(x=9.5\) and \(x=17\)

Determine the following probabilities for the standard normal distribution. a. \(P(-1.83 \leq z \leq 2.57)\) b. \(P(0 \leq z \leq 2.02)\) c. \(P(-1.99 \leq z \leq 0)\) d. \(P(z \geq 1.48)\)

A company that has a large number of supermarket grocery stores claims that customers who pay by personal checks spend an average of $$\$ 87$$ on groceries at these stores with a standard deviation of $$\$ 22 .$$ Assume that the expenses incurred on groceries by all such customers at these stores are normally distributed. a. Find the probability that a randomly selected customer who pays by check spends more than $$\$ 114$$ on groceries. b. What percentage of customers paying by check spend between $$\$ 40$$ and $$\$ 60$$ on groceries? c. What percentage of customers paying by check spend between $$\$ 70$$ and $$\$ 105 ?$$ d. Is it possible for a customer paying by check to spend more than $$\$ 185$$ ? Explain.
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