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Suppose a simple random sample of size \(n=36\) is obtained from a population with \(\mu=64\) and \(\sigma=18\) (a) Describe the sampling distribution of \(\bar{x}\). (b) What is \(P(\bar{x}<62.6) ?\) (c) What is \(P(\bar{x} \geq 68.7) ?\) (d) What is \(P(59.8<\bar{x}<65.9) ?\)

Short Answer

Expert verified
(a) Normal distribution with \(\mu = 64\) and \(SE = 3\). (b) \(0.3192\). (c) \(0.0590\). (d) \(0.6561\).

Step by step solution

01

Describe the Sampling Distribution

The sampling distribution of the sample mean \(\bar{x}\) will be normally distributed if the sample size is large enough (according to the Central Limit Theorem). Given \(\bar{x}\) is from a population with mean \(\mu = 64\) and standard deviation \(\sigma = 18\), the mean of the sampling distribution of \(\bar{x}\) is also \(\mu = 64\). The standard error (SE) of \(\bar{x}\) is calculated by SE = \(\sigma / \sqrt{n}\). Here, \(\sigma = 18\) and \(n = 36\), so SE = \(\frac{18}{\sqrt{36}} = 3\).
02

Find P(\bar{x}

To find \(P(\bar{x} < 62.6)\), first calculate the Z-score: \[Z = \frac{62.6 - \mu}{SE} = \frac{62.6 - 64}{3} = -0.467\] Use the Z-table, calculator, or an online tool to find the probability corresponding to Z = -0.467. \[P(\bar{x} < 62.6) \approx 0.3192 \]
03

Find P(\bar{x} \geq 68.7)

To find \(P(\bar{x} \geq 68.7)\), first calculate the Z-score: \[Z = \frac{68.7 - \mu}{SE} = \frac{68.7 - 64}{3} = 1.567\] Use the Z-table, calculator, or an online tool to find the probability. The Z-score for 1.567 corresponds to \[P(\bar{x} < 68.7) \approx 0.9410 \] Thus, \[P(\bar{x} \geq 68.7) = 1 - 0.9410 = 0.0590 \text{(or 5.9%)} \]
04

Find P(59.8

To find \(P(59.8 < \bar{x} < 65.9)\), calculate the Z-scores for both bounds. For 59.8: \[Z = \frac{59.8 - \mu}{SE} = \frac{59.8 - 64}{3} = -1.400\] For 65.9: \[Z = \frac{65.9 - 64}{SE} = \frac{65.9 - 64}{3} = 0.633\] Use a Z-table or tool to find corresponding probabilities. \[P(\bar{x} < 59.8) \approx 0.0808\] \[P(\bar{x} < 65.9) \approx 0.7369\] Thus, \[P(59.8 < \bar{x} < 65.9) = 0.7369 - 0.0808 = 0.6561 \]

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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 a fundamental principle in statistics. It states that the sampling distribution of the sample mean \( \bar{x} \) will approximate a normal distribution, regardless of the shape of the population distribution, provided the sample size is sufficiently large (typically \( n \geq 30\)). This is crucial because it allows us to make inferences about the sample mean using the properties of the normal distribution. In this exercise, since a sample size of \(n = 36\) is used, the CLT ensures that \( \bar{x} \) is approximately normally distributed with a mean (\( \mu \)) of 64 and a standard error (SE) of \(3\). This theorem helps simplify complex real-world data analysis into manageable normal distribution problems.
Z-score
A Z-score measures how many standard deviations an element is from the mean. It helps compare data points from different distributions. The formula to calculate the Z-score is: \[Z = \frac{X - \bar{x}}{\text{SE}} \]
In our exercise, the Z-scores are calculated to find the probabilities associated with different sample means. To find \( P(\bar{x} < 62.6) \), we calculated the Z-score as: \[Z = \frac{62.6 - 64}{3} = -0.467\] Similarly, for \( P(\bar{x} \geq 68.7)\) and \( P(59.8 < \bar{x} < 65.9)\), the Z-scores provide the probability positions. A Z-table or statistical software can then be used to determine these probabilities accurately. Understanding Z-scores allows for easier interpretation and comparison when dealing with normal distributions.
Standard Error
The standard error (SE) represents the standard deviation of the sampling distribution of a statistic, most commonly the mean. It indicates how much variability one can expect in the sample mean from sample to sample. SE is crucial when making inferences about the population mean from sample data.
The formula for SE is given as: \[ \text{SE} = \frac{\text{Population standard deviation (\ \sigma)}}{\text{Square root of the sample size (\ \sqrt{n} )}} \]
In our exercise, we use \( \sigma = 18\) and \( n = 36\): \[ \text{SE} = \frac{18}{\frac{\text{Square root of 36}}{}} = 3\]
This SE value then helps to calculate Z-scores and corresponding probabilities, thus quantifying the uncertainty in the estimation of the population mean. A smaller SE suggests that the sample mean is a more accurate reflection of the true population mean.

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Most popular questions from this chapter

The quality-control manager of a Long John Silver's restaurant wishes to analyze the length of time a car spends at the drive-through window waiting for an order. According to records obtained from the restaurants, it is determined that the mean time spent at the window is 59.3 seconds with a standard deviation of 13.1 seconds. The distribution of time at the window is skewed right (data based on information provided by Danica Williams, student at Joliet Junior College). (a) To obtain probabilities regarding a sample mean using the normal model, what size sample is required? (b) The quality-control manager wishes to use a new delivery system designed to get cars through the drivethrough system faster. A random sample of 40 cars results in a sample mean time spent at the window of 56.8 seconds. What is the probability of obtaining a sample mean of 56.8 seconds or less assuming the population mean is 59.3 seconds? Do you think that the new system is effective?

Suppose a simple random sample of size \(n=12\) is obtained from a population with \(\mu=64\) and \(\sigma=17\) (a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities regarding the sample mean? Assuming this condition is true, describe the sampling distribution of \(\bar{x}\) (b) Assuming the requirements described in part (a) are satisfied, determine \(P(\bar{x}<67.3)\) (c) Assuming the requirements described in part (a) are satisfied, determine \(P(\bar{x} \geq 65.2)\)

As the sample size increases, the difference between the sample mean, \(\bar{x},\) and the population mean, \(\mu,\) approaches ________.

Suppose a simple random sample of size \(n=75\) is obtained from a population whose size is \(N=10,000\) and whose population proportion with a specified characteristic is \(p=0.8\) (a) Describe the sampling distribution of \(\hat{p}\) (b) What is the probability of obtaining \(x=63\) or more individuals with the characteristic? That is, what is \(P(\hat{p} \geq 0.84) ?\) (c) What is the probability of obtaining \(x=51\) or fewer individuals with the characteristic? That is, what is \(P(\hat{p} \leq 0.68) ?\)

A nationwide study in 2003 indicated that about \(60 \%\) of college students with cell phones send and receive text messages with their phones. Suppose a simple random sample of \(n=1136\) college students with cell phones is obtained. (Source: promomagazine.com) (a) Describe the sampling distribution of \(\hat{p},\) the sample proportion of college students with cell phones who send or receive text messages with their phones. (b) What is the probability that 665 or fewer college students in the sample send and receive text messages with their cell phones? Is this result unusual? (c) What is the probability that 725 or more college students in the sample send and receive text messages with their cell phone? Is this result unusual?

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