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Chapter 6: Problem 26

How many close friends do you have? An opinion poll asks this question of an SRS of 1100 adults. Suppose that the number of close friends adults claim to have varies from person to person with mean \(\mu=9\) and standard deviation \(\sigma=2.5\) We will see later that in repeated random samples of size 1100 , the mean response \(\overline{x}\) will vary according to the Normal distribution with mean 9 and standard deviation 0.075 . What is \(P(8.9 \leq \overline{x} \leq 9.1)\) , the probability that the sample result \(\overline{x}\) estimates the population truth \(\mu=9\) to within \(\pm 0.1 ?\)

Short Answer

Expert verified
The probability that the sample mean is within \( \pm 0.1 \) of the population mean is 0.8164.

Step by step solution

01

Understand the Problem

We are asked to find the probability that the sample mean \( \overline{x} \) is between 8.9 and 9.1 for a sample size of 1100, given that the true population mean \( \mu \) is 9 and the standard deviation of the sample mean \( \sigma_{\overline{x}} \) is 0.075.
02

Standardize the Values

To find \( P(8.9 \leq \overline{x} \leq 9.1) \), we convert these \( \overline{x} \) values to z-scores using the formula: \[ z = \frac{\overline{x} - \mu}{\sigma_{\overline{x}}} \]For \( \overline{x} = 8.9 \), \[ z = \frac{8.9 - 9}{0.075} = -1.33 \] For \( \overline{x} = 9.1 \), \[ z = \frac{9.1 - 9}{0.075} = 1.33 \]
03

Use the Standard Normal Distribution

We'll use the standard normal distribution tables or a calculator to find the probability for the z-scores we calculated.The probability \( P(Z \leq 1.33) \) can be found in the z-table or using a calculator (approximately 0.9082), and \( P(Z \leq -1.33) \) is approximately 0.0918.
04

Calculate the Probability

The probability that 8.9 \( \leq \overline{x} \leq \) 9.1 is calculated using the difference between these probabilities:\[ P(8.9 \leq \overline{x} \leq 9.1) = P(Z \leq 1.33) - P(Z \leq -1.33) = 0.9082 - 0.0918 = 0.8164 \]

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Key Concepts

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

Normal Distribution
The normal distribution, often referred to as a bell curve, is a vital concept in statistics and probability. It describes how the values of a variable, most commonly found in various natural and social phenomena, are spread across different measurements. In a normal distribution:
  • Data has a symmetric distribution, meaning it is mirrored on both sides of its mean.
  • The mean, median, and mode of the data set are equal.
  • Approximately 68% of the data falls within one standard deviation from the mean, about 95% lies within two standard deviations, and around 99.7% falls within three standard deviations.
In our exercise, the problem states that the sample mean \((\overline{x})\) is normally distributed. This allows us to use the properties of the normal distribution to calculate probabilities, making use of z-scores with a standard deviation specific to the sample.
Sample Mean
The sample mean \((\overline{x})\) is an average of the observations in a sample. It's calculated by summing all measurements and dividing by the number of observations. This concept is crucial because it allows us to make inferences about a population based on a smaller group or sample of data.
  • The sample mean is a point estimate of the population mean \(\mu\).
  • In large samples, especially those that are normally distributed, the sample mean is a reliable estimation of the population mean.
  • In our example, the sample mean of adults' claimed number of close friends is used to estimate the true mean of all adults' number of close friends.
By having a large sample size of 1100, the law of large numbers ensures that our sample mean will closely approach the true population mean.
Z-scores
Z-scores are a statistical measurement that describe how many standard deviations a data point is from the mean. They are calculated using the formula:\[ z = \frac{(x - \mu)}{\sigma} \]This converts a normal distribution to a standard normal distribution (mean of 0 and standard deviation of 1), allowing us to make direct probabilistic assessments using Z-tables.
  • A z-score of 0 indicates that the data point's score is identical to the mean.
  • Positive z-scores indicate values above the mean, while negative scores indicate values below the mean.
  • In the exercise, we converted sample mean limits of 8.9 and 9.1 into z-scores to find probabilities from the standard normal distribution.
This conversion helps us calculate the probability of data points occurring within a certain range.
Standard Deviation
Standard deviation is a measure of how spread out the values in a data set are around the mean. It's important because it gives us insight into the variability present in the data. The formula for standard deviation is:\[ \sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}} \]Here, \(x_i\) represents each data point, \(\mu\) is the mean, and \(N\) is the number of data points.
  • A small standard deviation indicates that the data points are close to the mean, suggesting low variability.
  • A larger standard deviation means a wider spread of data points around the mean.
  • In the problem, the standard deviation of the sample mean, which is 0.075, is used to understand how much the sample mean is expected to vary from the population mean.
This helps in estimating probabilities concerning how close the sample mean may be to the actual population mean.

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Most popular questions from this chapter

Auto emissions The amount of nitrogen oxides \((\mathrm{NOX})\) present in the exhaust of a particular type of car varies from car to car according to a Normal distribution with mean 1.4 grams per mile (g/mi) and standard deviation 0.3 \(\mathrm{g} / \mathrm{mi}\) . Two randomly selected cars of this type are tested. One has 1.1 \(\mathrm{g} /\) mi of NOX; the other has 1.9 \(\mathrm{g} / \mathrm{mi}\) . The test station attendant finds this difference in emissions between two similar cars surprising. If the NOXlevels for two randomly chosen cars of this type are independent, find the probability that the difference is at least as large as the value the attendant observed. Follow the four-step process.

A balanced scale You have two scales for measuring weights in a chemistry lab. Both scales give answers that vary a bit in repeated weighings of the same item. If the true weight of a compound is 2.00 grams (g), the first scale produces readings \(X\) that have mean 2.000 g and standard deviation 0.002 g. The second scale's readings Y have mean 2.001 \(\mathrm{g}\) and standard deviation 0.001 \(\mathrm{g} .\) The readings \(X\) and \(\mathrm{Y}\) are independent. Find the mean and standard deviation of the difference \(Y-\mathrm{X}\) between the readings. Interpret each value in context.

Statistics for investing \((3.2)\) Joe's retirement plan invests in stocks through an "index fund" that follows the behavior of the stock market as a whole, as measured by the Standard \& Poor's (S\&P) 500 stock index. Joe wants to buy a mutual fund that does not track the index closely. He reads that monthly returns from Fidelity Technology Fund have correlation \(r=\) 0.77 with the S\&P 500 index and that Fidelity Real Estate Fund has correlation \(r=0.37\) with the index. (a) Which of these funds has the closer relationship to returns from the stock market as a whole? How do you know? (b) Does the information given tell Joe anything about which fund had higher returns?

Cranky mower To start her old mower, Rita has to pull a cord and hope for some luck. On any particular pull, the mower has a 20\(\%\) chance of starting. (a) Find the probability that it takes her exactly 3 pulls to start the mower. Show your work. (b) Find the probability that it takes her more than 10 pulls to start the mower. Show your work.

A sample survey contacted an SRS of 663 registered voters in Oregon shortly after an clection and asked respondents whether they had voted. Voter records show that 56\(\%\) of registered voters had actually voted. We will see later that in repeated random samples of size 663 , the proportion in the sample who voted (call this proportion With vary according to the Normal distribution with mean \(\mu=0.56\) and standard deviation \(\sigma=0.019\) (a) If the respondents answer truthfully, what is \(\mathrm{P}(0.52 \leq V \leq 0.60)\) ? This is the probability that the sample proportion \(V\) estimates the population proportion 0.56 within \(\pm 0.04\) (b) In fact, 72\(\%\) of the respondents said they had voted \((V=0.72) .\) If respondents answer truthfully, what is \(P(V \geq 0.72) ?\) This probability is so small that it is good evidence that some people who did not vote claimed that they did vote.
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