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Chapter 7: Problem 47

Let \(x\) be a continuous random variable that follows a distribution skewed to the left with \(\mu=90\) and \(\sigma=18\). Assuming \(n / N \leq .05\), find the probability that the sample mean, \(\bar{x}\), for a random sample of 64 taken from this population will be a. less than \(82.3\) b. greater than \(86.7\)

Short Answer

Expert verified
The probability that the sample mean will be less than 82.3 is 0.0003 and greater than 86.7 is 0.9292.

Step by step solution

01

Calculate the standard error

Firstly, we need to calculate the standard error which measures the dispersion of sample means. This is given by \( \sigma / \sqrt{n} \) where \( \sigma \) is the standard deviation and \( n \) is the sample size. In this case, \( \sigma = 18 \) and \( n = 64 \). So, standard error = \( 18 / \sqrt{64} = 2.25 \).
02

Calculate the Z scores

The next step is to convert the sample means to Z scores, which is calculated by \( (x - \mu) / \text{standard error} \). For \( x = 82.3 \), the Z score = \( (82.3 - 90) / 2.25 = -3.42 \) and for \( x = 86.7 \), the Z score = \( (86.7 - 90) / 2.25 = -1.47 \).
03

Find the probability using Z table

Once we have the Z scores, we use a Z-table to find the corresponding probabilities. Referring to the Z-table, the probability corresponding to Z = -3.42 (for x < 82.3) is 0.0003 and for Z = -1.47 (for x > 86.7) is 0.0708. However, we need the probability for x > 86.7 which would be 1 - 0.0708 = 0.9292.

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Most popular questions from this chapter

A Census Bureau report revealed that \(43.7 \%\) of Americans who moved between 2009 and 2010 did so for housing-related reasons, such as the desire to live in a new or better home or apartment (http://www.census.gov/newsroom/releases/archives/mobility_of_the_population/cb11-91.html). Suppose that this percentage is true for the current population of Americans. a. Suppose that \(49 \%\) of the people in a random sample of 100 Americans who moved recently did so for housing-related reasons. How likely is it for the sample proportion in a sample of 100 to be \(.49\) or more when the population proportion is \(.437 ?\) b. Refer to part a. How likely is it for the sample proportion in a random sample of 200 to be 49 or more when the population proportion is .437? c. What is the smallest sample size that will produce a sample proportion of \(.49\) or more in no more than \(5 \%\) of all sample surveys of that size?

According to a report in The New York Times, bank tellers in the United States earn an average of $$\$ 25,510$$ a year (Jessica Silver-Greenberg, The New York Times, April 22, 2012). Suppose that the current distribution of salaries of all bank tellers in the United States has a mean of $$\$ 25,510$$ and a standard deviation of $$\$ 4550 .$$ Let \(\bar{x}\) be the average salary of a random sample of 200 such tellers. Find the mean and standard deviation of the sampling distribution of \(\bar{x}\).

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Let \(x\) be a continuous random variable that has a distribution skewed to the right with \(\mu=60\) and \(\sigma=10\). Assuming \(n / N \leq .05\), find the probability that the sample mean, \(\bar{x}\), for a random sample of 40 taken from this population will be a. less than \(62.20\) b. between \(61.4\) and \(64.2\)

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