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Suppose a random sample of \(n=25\) observations is selected from a population that is normally distributed with mean equal to 106 and standard deviation equal to 12 a. Give the mean and the standard deviation of the sampling distribution of the sample mean \(\bar{x}\). b. Find the probability that \(\bar{x}\) exceeds \(110 .\) c. Find the probability that the sample mean deviates from the population mean \(\mu=106\) by no more than 4

Short Answer

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Question: A random sample of 25 observations is selected from a normally distributed population with a known mean of 106 and a standard deviation of 12. Compute the probabilities of the following events: a) The sample mean deviates from the population mean by no more than 4. b) The sample mean exceeds 110. Answer: The probability that the sample mean deviates from the population mean by no more than 4 is 0.905. The probability that the sample mean exceeds 110 is 0.0475.

Step by step solution

01

Find the Mean of the Sampling Distribution of The Sample Mean

Given that the population mean is 106, the mean of the sampling distribution of the sample mean will also be 106. This is because the sample mean is an unbiased estimator of the population mean.
02

Find the Standard Deviation of the Sampling Distribution of the Sample Mean

Given that the population standard deviation is 12 and the sample size is 25, we can find the standard deviation of the sampling distribution of the sample mean using the formula: \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\) Where \(\sigma_{\bar{x}}\) is the standard deviation of the sample mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Plug in the given values: \(\sigma_{\bar{x}} = \frac{12}{\sqrt{25}}= \frac{12}{5}\) So, the standard deviation of the sampling distribution of the sample mean is 2.4.
03

Find the Z-Scores Corresponding to the Given Probabilities

To find the probability that the sample mean deviates from the population mean by no more than 4 and the probability that it exceeds 110, we need to find the respective Z-scores. The Z-score is calculated as follows: \(Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}\) To find the probability that \(\bar{x}\) exceeds 110, plug the values into the formula: \(Z = \frac{110 - 106}{2.4} = \frac{4}{2.4} ≈ 1.67\) To find the probability that the sample mean deviates from the population mean by no more than 4, we need to find the Z-scores for the range [102, 110]. For 102, the Z-score is: \(Z = \frac{102 - 106}{2.4} = \frac{-4}{2.4} ≈ -1.67\)
04

Find the Probabilities Using a Standard Normal Table

Using a standard normal table or calculator, we can find the probabilities for each Z-score. For \(Z = 1.67\), the probability is 0.9525. The probability that the sample mean exceeds 110 is found by subtracting this probability from 1: \(P(\bar{x}>110) = 1 - 0.9525 = 0.0475\) Next, to find the probability that the sample mean deviates from the population mean by no more than 4, we calculate the probability for the range of Z-scores between -1.67 and 1.67. The probability for \(Z = -1.67\) is 0.0475. Subtracting the lower Z-score probability from the higher Z-score probability gives: \(P(102\le\bar{x}\le110) = 0.9525 - 0.0475 = 0.905\)
05

Summarize the Results

We found the mean and standard deviation of the sampling distribution of the sample mean to be 106 and 2.4, respectively. The probability that the sample mean exceeds 110 is 0.0475, and the probability that the sample mean deviates from the population mean by no more than 4 is 0.905.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem
The Central Limit Theorem (CLT) is one of the cornerstones of statistics. It tells us that when we take a sufficiently large sample size from a population with any distribution, the sampling distribution of the sample mean will approximate a normal distribution. Even if the original population is not normally distributed, the sample mean's distribution will be normal, as long as the sample size is large enough.

In our exercise, the sample size is 25, which is often considered adequate to apply the CLT. However, since the population is known to be normally distributed from the start, the CLT assures us that the sample mean distribution is normal too, without the need for a large sample size.
  • This distribution of the sample mean will center around the population mean.
  • The standard deviation of this distribution will be smaller than that of the population, specifically using the formula \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\).
Understanding the CLT helps us assess probabilities about sample means and their deviations from the theoretical population mean. It is foundational for performing hypothesis testing and constructing confidence intervals.
Z-Score
The Z-Score is a statistical measurement that describes a value's relation to the mean of a group of values. In terms of the sampling distribution, the Z-score helps us determine how far and in what direction a sample mean deviates from the population mean, measured in terms of standard deviations.

To calculate the Z-score for our sample mean, we use the formula:
  • \( Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \)
The numerator represents the deviation of the sample mean from the population mean, while the denominator standardizes this deviation by the sampling distribution's standard deviation.

For example, in our exercise, to find out if the sample mean exceeds 110, we calculate the Z-score as follows:
  • \( Z = \frac{110 - 106}{2.4} \approx 1.67 \)
This Z-score represents the number of standard deviations the score of 110 is from the population mean of 106. Using a Z-score, we can consult a standard normal table to find the probability associated with this deviation.
Normal Distribution
The normal distribution, often referred to as the "bell curve," is a probability distribution that is symmetric about the mean. Most values cluster around a central peak, and probabilities for values taper off equally on both sides.

Normal distributions are vital in statistics because they represent many natural phenomena. For the purposes of inferential statistics, such as calculating probabilities from sample data, assuming a normal distribution simplifies calculations and interpretations.
  • The mean, median, and mode of a normal distribution are all equal.
  • The curve is symmetric around the mean.
In the scenario we're analyzing, the sampling distribution of our sample mean is normal due to the known normality of the population and the CLT. This allows us to calculate probabilities and make predictions about the sample mean, using the Z-score and normal tables.

For instance, once the sampling distribution is known to be normal, it is straightforward to calculate the probability that the mean deviates from 106 by no more than 4. This involves considering the range from 102 to 110, translating this range into Z-scores, and using the properties of the normal distribution to find probabilities.

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Most popular questions from this chapter

News reports tell us that the average American is overweight. Many of us have tried to trim down to our weight when we finished high school or college. And, in fact, only \(19 \%\) of adults say they do not suffer from weight-loss woes. Suppose that the \(19 \%\) figure is correct, and that a random sample of \(n=100\) adults is selected. a. Does the distribution of \(\hat{p},\) the sample proportion of adults who do not suffer from excess weight, have an approximate normal distribution? If so, what is its mean and standard deviation? b. What is the probability that the sample proportion \(\hat{p}\) exceeds .25? c. What is the probability that \(\hat{p}\) lies within the interval .25 to \(.30 ?\) d. What might you conclude about \(p\) if the sample proportion exceeded .30?

A manufacturer of paper used for packaging requires a minimum strength of 20 pounds per square inch. To check on the quality of the paper, a random sample of 10 pieces of paper is selected each hour from the previous hour's production and a strength measurement is recorded for each. The standard deviation \(\sigma\) of the strength measurements, computed by pooling the sum of squares of deviations of many samples, is known to equal 2 pounds per square inch, and the strength measurements are normally distributed. a. What is the approximate sampling distribution of the sample mean of \(n=10\) test pieces of paper? b. If the mean of the population of strength measurements is 21 pounds per square inch, what is the approximate probability that, for a random sample of \(n=10\) test pieces of paper, \(\bar{x}<20 ?\) c. What value would you select for the mean paper strength \(\mu\) in order that \(P(\bar{x}<20)\) be equal to \(.001 ?\)

Random samples of size \(n=75\) were selected from a binomial population with \(p=.4 .\) Use the normal distribution to approximate the following probabilities: a. \(P(\hat{p} \leq .43)\) b. \(P(.35 \leq \hat{p} \leq .43)\)

A bottler of soft drinks packages cans in six-packs. Suppose that the fill per can has an approximate normal distribution with a mean of 12 fluid ounces and a standard deviation of 0.2 fluid ounces. a. What is the distribution of the total fill for a case of 24 cans? b. What is the probability that the total fill for a case is less than 286 fluid ounces? c. If a six-pack of soda can be considered a random sample of size \(n=6\) from the population, what is the probability that the average fill per can for a six-pack of soda is less than 11.8 fluid ounces?

Some sports that involve a significant amount of running, jumping, or hopping put participants at risk for Achilles tendinopathy (AT), an inflammation and thickening of the Achilles tendon. A study in The American Journal of Sports Medicine looked at the diameter (in \(\mathrm{mm}\) ) of the affected and nonaffected tendons for patients who participated in these types of sports activities. \(^{9}\) Suppose that the Achilles tendon diameters in the general population have a mean of 5.97 millimeters (mm) with a standard deviation of \(1.95 \mathrm{~mm} .\) a. What is the probability that a randomly selected sample of 31 patients would produce an average diameter of \(6.5 \mathrm{~mm}\) or less for the nonaffected tendon? b. When the diameters of the affected tendon were measured for a sample of 31 patients, the average diameter was \(9.80 .\) If the average tendon diameter in the population of patients with AT is no different than the average diameter of the nonaffected tendons \((5.97 \mathrm{~mm})\), what is the probability of observing an average diameter of 9.80 or higher? c. What conclusions might you draw from the results of part b?

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