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A population has a mean of 200 and a standard deviation of \(50 .\) Suppose a simple random sample of size 100 is selected and \(\bar{x}\) is used to estimate \(\mu\) a. What is the probability that the sample mean will be within ±5 of the population mean? b. What is the probability that the sample mean will be within ±10 of the population mean?

Short Answer

Expert verified
a. 0.6826 b. 0.9544

Step by step solution

01

Understanding the Problem

We are given a population with a mean \(\mu = 200\) and a standard deviation \(\sigma = 50\). We need to find the probability that the sample mean \(\bar{x}\) of a sample of size \(n = 100\) is within \(\pm 5\) and \(\pm 10\) of the population mean. This requires us to use the Central Limit Theorem and calculate probabilities using the standard normal distribution.
02

Calculate Standard Error of the Mean

The standard error of the mean (SEM) is given by the formula: \[ SEM = \frac{\sigma}{\sqrt{n}} \] Here, \(\sigma = 50\) and \(n = 100\). Therefore, \[ SEM = \frac{50}{\sqrt{100}} = 5 \]
03

Find Z-scores for ±5 Interval

The Z-score formula is: \[ Z = \frac{X - \mu}{SEM} \] For ±5 interval, the limits are 195 and 205. The Z-scores are: \[ Z_1 = \frac{195 - 200}{5} = -1 \] \[ Z_2 = \frac{205 - 200}{5} = 1 \]
04

Calculate Probability for ±5 Interval

Using the standard normal table, we find the probability for \(Z = -1\) to \(Z = 1\). \[ P(-1 < Z < 1) = P(Z < 1) - P(Z < -1) \]Looking at the Z-table: \[ P(Z < 1) = 0.8413 \]\[ P(Z < -1) = 0.1587 \]\[ P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826 \]
05

Find Z-scores for ±10 Interval

For ±10 interval, the limits are 190 and 210. The Z-scores are: \[ Z_3 = \frac{190 - 200}{5} = -2 \] \[ Z_4 = \frac{210 - 200}{5} = 2 \]
06

Calculate Probability for ±10 Interval

Using the standard normal table, we find the probability for \(Z = -2\) to \(Z = 2\). \[ P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) \]Looking at the Z-table:\[ P(Z < 2) = 0.9772 \]\[ P(Z < -2) = 0.0228 \]\[ P(-2 < Z < 2) = 0.9772 - 0.0228 = 0.9544 \]

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Key Concepts

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

Standard Error of the Mean
The Standard Error of the Mean (SEM) is a fundamental concept in statistics that helps us understand how much sample means might vary if we were to repeatedly draw samples from the same population. This is crucial when we don't observe the entire population, but rather a random sample. The SEM provides a measure of the expected disparity between the sample mean and the true population mean.
The formula to calculate the standard error is given by:
  • \(SEM = \frac{\sigma}{\sqrt{n}}\)
where \(\sigma\) is the population standard deviation, and \(n\) is the sample size. In the context of our exercise, the SEM is determined with \(\sigma = 50\) and \(n = 100\).
Calculating, we find that:
  • \(SEM = \frac{50}{\sqrt{100}} = 5\)
This SEM value tells us how much variability we can expect in the sample means from one sample to another, offering us a foundation for further probability calculations.
Z-score
The Z-score is a critical statistic that indicates how many standard deviations an element is from the mean. When analyzing sample data, Z-scores allow us to understand the likelihood of a sample mean occurring within a certain range, relative to the population mean. A Z-score tells us where a particular score lies compared to the norm, making it an essential tool in hypothesis testing and confidence intervals.
To calculate a Z-score, we use the formula:
  • \(Z = \frac{X - \mu}{SEM}\)
where \(X\) is the value we are comparing, \(\mu\) is the population mean, and \(SEM\) is the standard error of the mean previously calculated.
In our exercise:
  • For a ±5 interval around the mean, the limits are 195 and 205, producing Z-scores of -1 and 1 respectively.
  • For a ±10 interval around the mean, the limits are 190 and 210, producing Z-scores of -2 and 2 respectively.
These scores help us find the probabilities associated with observing such sample means, particularly when converted to probabilities using the standard normal distribution.
Standard Normal Distribution
The Standard Normal Distribution is a special type of normal distribution where the mean is 0 and the standard deviation is 1. This distribution is crucial because it allows us to use the Z-scores to calculate probabilities for events occurring within a certain range.
The beauty of the standard normal distribution lies in its universality; it standardizes different datasets, enabling us to compare them regardless of their original scales. Using a standard normal distribution table, or Z-table, we cross-reference Z-scores to find probabilities.
For instance, our exercise involves finding probabilities within certain Z-score boundaries:
  • For the ±5 range, the Z-scores of -1 and 1 correspond to probability calculations where the total probability between these two values is approximately 0.6826.
  • For the ±10 range, the Z-scores of -2 and 2 extend this range, leading to a broader probability coverage of approximately 0.9544.
These probabilities represent the likelihood that a randomly selected sample mean will fall within these boundaries, illustrating the robustness of the Standard Normal Distribution in statistical analysis.

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Most popular questions from this chapter

Americans have become increasingly concerned about the rising cost of Medicare. In 1990 the average annual Medicare spending per enrollee was \(\$ 3267 ;\) in \(2003,\) the average annual Medicare spending per enrollee was \(\$ 6883\) (Money, Fall 2003 ). Suppose you hired a consulting firm to take a sample of fifty 2003 Medicare enrollees to further investigate the nature of expenditures. Assume the population standard deviation for 2003 was \(\$ 2000\). a. Show the sampling distribution of the mean amount of Medicare spending for a sample of fifty 2003 enrollees. b. What is the probability that the sample mean will be within \(\pm \$ 300\) of the population mean? c. What is the probability that the sample mean will be greater than \(\$ 7500 ?\) If the consulting firm tells you the sample mean for the Medicare enrollees it interviewed was \(\$ 7500,\) would you question whether the firm followed correct simple random sampling procedures? Why or why not?

A sample of 50 Fortune 500 companies (Fortune, April 14,2003 ) showed 5 were based in New York, 6 in California, 2 in Minnesota, and 1 in Wisconsin. a. Develop an estimate of the proportion of Fortune 500 companies based in New York. b. Develop an estimate of the number of Fortune 500 companies based in Minnesota. c. Develop an estimate of the proportion of Fortune 500 companies that are not based in these four states.

Indicate which of the following situations involve sampling from a finite population and which involve sampling from an infinite population. In cases where the sampled population is finite, describe how you would construct a frame. a. \(\quad\) Obtain a sample of licensed drivers in the state of New York. b. Obtain a sample of boxes of cereal produced by the Breakfast Choice company. c. Obtain a sample of cars crossing the Golden Gate Bridge on a typical weekday. d. Obtain a sample of students in a statistics course at Indiana University. e. Obtain a sample of the orders that are processed by a mail-order firm.

The average score for male golfers is 95 and the average score for female golfers is 106 (Golf Digest, April 2006). Use these values as the population means for men and women and assume that the population standard deviation is \(\sigma=14\) strokes for both. A simple random sample of 30 male golfers and another simple random sample of 45 female golfers will be taken. a. Show the sampling distribution of \(\bar{x}\) for male golfers. b. What is the probability that the sample mean is within three strokes of the population mean for the sample of male golfers? c. What is the probability that the sample mean is within three strokes of the population mean for the sample of female golfers? d. In which case, part (b) or part (c), is the probability of obtaining a sample mean within three strokes of the population mean higher? Why?

In 2008 the Better Business Bureau settled \(75 \%\) of complaints it received (USA Today, March 2,2009 ). Suppose you have been hired by the Better Business Bureau to investigate the complaints it received this year involving new car dealers. You plan to select a sample of new car dealer complaints to estimate the proportion of complaints the Better Business Bureau is able to settle. Assume the population proportion of complaints settled for new car dealers is \(.75,\) the same as the overall proportion of complaints settled in 2008 a. Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of \(\bar{p}\) b. Based upon a sample of 450 complaints, what is the probability that the sample proportion will be within .04 of the population proportion? c. Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of \(\bar{p}\) d. Based upon the smaller sample of only 200 complaints, what is the probability that the sample proportion will be within .04 of the population proportion? e. As measured by the increase in probability, how much do you gain in precision by taking the larger sample in part (b)?

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