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Chapter 7: Problem 62

Suppose a random sample of \(n=5\) observations is selected from a population that is normally distributed, with mean equal to 1 and standard deviation equal to \(.36 .\) a. Give the mean and standard deviation of the sampling distribution of \(\bar{x}\). b. Find the probability \(\bar{x}\) that exceeds \(1.3 .\) c. Find the probability that the sample mean \(\bar{x}\) is less than .5 . d. Find the probability that the sample mean deviates from the population mean \(\mu=1\) by more than .4 .

Short Answer

Expert verified
Additionally, determine the probabilities of the sample mean (a) exceeding 1.3, (b) being less than 0.5, and (c) deviating from the population mean by more than 0.4. Answer: The mean of the sampling distribution is 1, and the standard deviation is approximately 0.1612. The probabilities for each scenario are as follows: (a) approximately 0.0314, (b) approximately 0.0010, and (c) approximately 0.0132.

Step by step solution

01

Calculate the mean and standard deviation of the sampling distribution of the sample mean

The mean and standard deviation of the sampling distribution of the sample mean are given by \(\mu_{\bar{x}} = \mu\) and \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\). In our case, \(\mu = 1\) and \(\sigma = 0.36\). Therefore, for the given sample size \(n = 5\): $$ \mu_{\bar{x}} = \mu = 1 \\ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{0.36}{\sqrt{5}} \approx 0.1612 $$
02

Standardize \(\bar{x}\)

To find the probabilities for part (b), (c) and (d), we need to standardize the sample mean using the standard normal distribution with \(Z = \frac{\bar{x} - \mu_{\bar{x}} }{\sigma_{\bar{x}}}\). (a) Mean of the sampling distribution is \(1\), and standard deviation is approximately \(0.1612\).
03

Find the probability \(\bar{x}\) that exceeds \(1.3\)

To find this probability, we need to standardize the value of \(1.3\) using the previously computed mean and standard deviation of the sampling distribution: $$ Z = \frac{1.3-1}{0.1612} \approx 1.86 $$ Now, we need to calculate the probability that \(Z\) exceeds \(1.86\). Using the standard normal table or calculator, we find: $$ P(Z > 1.86) \approx 0.0314 $$ So, the probability that \(\bar{x}\) exceeds \(1.3\) is approximately \(0.0314\). (b) The probability that \(\bar{x}\) exceeds \(1.3\) is approximately \(0.0314\).
04

Find the probability that the sample mean \(\bar{x}\) is less than .5

Similarly, to find this probability, we need to standardize the value of \(0.5\): $$ Z = \frac{0.5-1}{0.1612} \approx -3.1012 $$ Now, we need to calculate the probability that \(Z\) is less than \(-3.1012\). Using the standard normal table or calculator, we find: $$ P(Z < -3.1012) \approx 0.0010 $$ So, the probability that \(\bar{x}\) is less than \(0.5\) is approximately \(0.0010\). (c) The probability that the sample mean \(\bar{x}\) is less than \(0.5\) is approximately \(0.0010\).
05

Find the probability that the sample mean deviates from the population mean by more than 0.4

To find this probability, we need to find the probability that \(\bar{x}\) is either more than \(1.4\) or less than \(0.6\). We first standardize these values: $$ Z_{0.6} = \frac{0.6-1}{0.1612} \approx -2.477 \\ Z_{1.4} = \frac{1.4-1}{0.1612} \approx 2.477 $$ Using the standard normal table or calculator, we find: $$ P(Z < -2.477) \approx 0.0066 \\ P(Z > 2.477) \approx 0.0066 $$ The probability that the sample mean deviates from the population mean by more than 0.4 is the sum of these probabilities: $$ P(\bar{x} < 0.6 \, \text{or} \, \bar{x} > 1.4) = P(Z < -2.477) + P(Z > 2.477) \approx 0.0066 + 0.0066 \approx 0.0132 $$ (d) The probability that the sample mean deviates from the population mean by more than 0.4 is approximately \(0.0132\).

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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 is a fundamental concept in statistics, often depicted as a bell-shaped curve. It's symmetrical around its mean, which means that the distribution on the left and right of the central peak are mirror images of each other. This type of distribution is handy because many natural phenomena follow a normal distribution, including heights, test scores, and measurement errors.
  • The mean (\( \mu \)) is the central point of the distribution.
  • The standard deviation (\( \sigma \)) measures the spread; larger values indicate a wider spread of data.

In a standard normal distribution, the mean is 0 and the standard deviation is 1. Observations can be transformed into a standard normal distribution using the Z-score formula: \( Z = \frac{x - \mu}{\sigma} \). This transformation allows us to calculate probabilities and percentiles for various scenarios.
Sample Mean
The sample mean (\( \bar{x} \)) is the average of observations in a sample. Calculating the sample mean is a basic step in statistics to summarize the data collected, giving a simple snapshot of what's typical in the sample.
  • It is calculated as \( \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \), where \( n \) is the number of observations.

When working with a sample mean, especially if the population is normally distributed, the sampling distribution of the sample mean will also be normal. This is particularly useful because it simplifies analysis thanks to the properties of normal distributions.
Standard Deviation
The standard deviation quantifies the amount of variation or dispersion in a set of data values. A low standard deviation indicates that the data points are close to the mean. Conversely, high standard deviation suggests a wider deviation from the mean.
  • For a population,\( \sigma = \sqrt{\frac{\sum_{i=1}^{N} (x_i - \mu)^2}{N}} \).
  • For a sample, \( s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}} \).

Understanding standard deviation helps in assessing how spread out the data is, and is essential in the calculation of the standard deviation of the sampling distribution: \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), particularly useful for probability calculations.
Probability Calculation
Probability calculations involve determining the likelihood of an event occurring based on a given probability distribution. In the context of the normal distribution and sampling, we often calculate probabilities relating to the sample mean.
  • Standardization (finding a Z-score): \( Z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \) allows us to find probabilities using standard normal distribution tables or software.
  • The calculated Z-score helps in determining how far away \( \bar{x} \) is from the mean (\( \mu_{\bar{x}} \)).

Once standardized, you can use tables or computational tools to find cumulative probabilities or tail areas, which reflect the probability of events occurring within or outside specified standards. This process is crucial in hypothesis testing, confidence intervals, and more nuanced statistical analyses.

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