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Chapter 7: Problem 36

The times that college students spend studying per week have a distribution that is skewed to the right with a mean of \(8.4\) hours and a standard deviation of \(2.7\) hours. Find the probability that the mean time spent studying per week for a random sample of 45 students would be a. between 8 and 9 hours b. less than 8 hours

Short Answer

Expert verified
The probability that the mean time spent studying per week for a random sample of 45 students would be a) between 8 and 9 hours is approximately \(0.7732\) or \(77.32\% \), and b) less than 8 hours is approximately \(0.1587\) or \(15.87\%\).

Step by step solution

01

Calculate the standard deviation of the sample mean

The standard error (SE) or standard deviation of the sample mean is calculated by dividing the population standard deviation \(蟽\) by the square root of the sample size \(n\). So, SE = \(蟽/鈭歯\). Here, \(蟽 = 2.7\) and \(n = 45\). This gives us SE = \(2.7/鈭�45 鈮� 0.402\).
02

Convert X values to Z score to find probability

The Z score is found by subtracting mean (渭) from value (X) and then divided by standard deviation (SE). So, \(Z = (X - 渭) / SE\). For both parts: a) \(Z_a (for between 8 and 9 hours) = (X - 渭) / SE = (8 - 8.4) / 0.402 鈮� -1\) and \(Z_b (for 9 hours) = (9 - 8.4) / 0.402 鈮� 1.49\). Hence, we need to find the area between -1 and 1.49, representing the probability. b) \(Z (for less than 8 hours) = (X - 渭) / SE = (8 - 8.4) / 0.402 鈮� -1\). Hence, we need to find the area to the left of -1.
03

Find the probabilities

We can find these probabilities using the standard normal distribution table or a calculator with a function for the cumulative density function. For part a, this probability is \(P(-1 < Z < 1.49) = P(Z < 1.49) - P(Z < -1) = 0.9319 - 0.1587 = 0.7732\). For part b, this probability is \(P(Z < -1) = 0.1587\).

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Most popular questions from this chapter

In a January 2014 survey conducted by the Associated PressWe TV, \(68 \%\) of American adults said that owning a home is the most important thing or \(a\) very important but not the most important thing (opportunityagenda.org). Assume that this percentage is true for the current population of American adults. Let \(\hat{p}\) be the proportion in a random sample of 1000 American adults who hold the above opinion. Find the mean and standard deviation of the sampling distribution of \(\hat{p}\) and describe its shape.

A machine at Katz Steel Corporation makes 3 -inch-long nails. The probability distribution of the lengths of these nails is approximately normal with a mean of 3 inches and a standard deviation of \(.1\) inch. The quality control inspector takes a sample of 25 nails once a week and calculates the mean length of these nails. If the mean of this sample is either less than \(2.95\) inches or greater than \(3.05\) inches, the inspector concludes that the machine needs an adjustment. What is the probability that based on a sample of 25 nails, the inspector will conclude that the machine needs an adjustment?

How does the value of \(\sigma_{\bar{x}}\) change as the sample size increases? Explain.

Briefly explain the meaning of a population probability distribution and a sampling distribution. Give an example of each.

Explain briefly the meaning of nonsampling errors. Give an example. Do such errors occur only in a sample survey, or can they occur in both a sample survey and a census?
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