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Chapter 6: Problem 7

a. Find a \(z_{0}\) such that \(P\left(z>z_{0}\right)=.025 .\) b. Find a \(z_{0}\) such that \(P\left(z

Short Answer

Expert verified
b) The probability of \(z\) being less than \(z_{0}\) is equal to 0.9251? a) \(z_{0} \approx 1.96\) b) \(z_{0} \approx 1.44\)

Step by step solution

01

Recall the standard normal distribution table

To solve this exercise, we will use the standard normal distribution table, also known as the Z-table, which provides us the probabilities associated with specific z-scores in a standard normal distribution. If you have access to a calculator with the inverse cumulative distribution function, you can use this as well.
02

Find \(z_{0}\) for part a

For part a, we want to find \(z_{0}\) such that \(P(z>z_{0}) = 0.025\). Since the normal distribution is symmetric, we can re-write this as \(P(z<-z_{0}) = 0.025\). Now, using the standard normal distribution table or a calculator, we look for the value of \(z_{0}\) where the probability of \(z\) being less than \(-z_{0}\) is 0.025. We find that this value is approximately \(z_{0} = 1.96\).
03

Find \(z_{0}\) for part b

For part b, we want to find \(z_{0}\) such that \(P(z<z_{0}) = 0.9251\). Using the standard normal distribution table or a calculator, we look for the value of \(z_{0}\) where the probability of \(z\) being less than \(z_{0}\) is 0.9251. We find that this value is approximately \(z_{0} = 1.44\).
04

Final Answers

a. \(z_{0} \approx 1.96\) b. \(z_{0} \approx 1.44\)

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Most popular questions from this chapter

A manufacturing plant uses 3000 electric light bulbs whose life spans are normally distributed, with mean and standard deviation equal to 500 and 50 hours, respectively. In order to minimize the number of bulbs that burn out during operating hours, all the bulbs are replaced after a given period of operation. How often should the bulbs be replaced if we wish no more than \(1 \%\) of the bulbs to burn out between replacement periods?

How often do you watch movies at home? A USA Today Snapshot found that about 7 in 10 adults say they watch movies at home at least once a week. \(^{5}\) Suppose a random sample of \(n=50\) adults are polled and asked if they had watched a movie at home this week. Let us assume that \(p=.7\) is, in fact, correct. What are the probabilities for the following events? a. Fewer than 30 individuals watched a movie at home this week? b. More than 42 individuals watched a movie at home this week? c. Fewer than 10 individuals did not watch a movie at home this week?

In Exercise 6.28 , we suggested that the IRS assign auditing rates per state by randomly selecting 50 auditing percentages from a normal distribution with a mean equal to \(1.55 \%\) and a standard deviation of \(.45 \%\) a. What is the probability that a particular state would have more than \(2 \%\) of its tax returns audited? b. What is the expected value of \(x\), the number of states that will have more than \(2 \%\) of their income tax returns audited? c. Is it likely that as many as 15 of the 50 states will have more than \(2 \%\) of their income tax returns audited?

Let \(x\) be a binomial random variable for \(n=25,\) \(p=.2\) a. Use Table 1 in Appendix I to calculate \(P(4 \leq x \leq 6)\). b. Find \(\mu\) and \(\sigma\) for the binomial probability distribution, and use the normal distribution to approximate the probability \(P(4 \leq x \leq 6)\). Note that this value is a good approximation to the exact value of \(P(4 \leq x \leq 6)\) even though \(n p=5\)

One method of arriving at economic forecasts is to use a consensus approach. A forecast is obtained from each of a large number of analysts, and the average of these individual forecasts is the consensus forecast. Suppose the individual 2008 January prime interest rate forecasts of economic analysts are approximately normally distributed with the mean equal to \(8.5 \%\) and a standard deviation equal to \(0.2 \% .\) If a single analyst is randomly selected from among this group, what is the probability that the analyst's forecast of the prime rate will take on these values? a. Exceed \(8.75 \%\) b. Be less than \(8.375 \%\)
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