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Chapter 7: Problem 5

A synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with mean 75.5 psi and standard deviation 3.5 psi. Find the probability that a random sample of \(n=6\) fiber specimens will have sample mean tensile strength that exceeds 75.75 psi.

Short Answer

Expert verified
The probability is approximately 0.431.

Step by step solution

01

Understand the Problem

We are given a normal distribution with a mean tensile strength of 75.5 psi and a standard deviation of 3.5 psi. We are asked to find the probability that the sample mean of 6 fiber specimens exceeds 75.75 psi.
02

Identify the Distribution of the Sample Mean

For a sample mean, we use the Central Limit Theorem, which states that the sampling distribution of the sample mean will also be normally distributed with the same mean as the original distribution and a standard deviation equal to the population standard deviation divided by the square root of the sample size. For this problem, the mean remains 75.5 psi, but the standard deviation becomes \(\frac{3.5}{\sqrt{6}}\).
03

Calculate the Standard Error

The standard error of the sample mean is \(\frac{3.5}{\sqrt{6}}\). Calculate this value to find the standard deviation of the sampling distribution: \(\frac{3.5}{\sqrt{6}} \approx 1.428\).
04

Find the Z-score

The Z-score represents how many standard deviations the desired value (75.75 psi in this problem) is from the mean of the distribution of the sample mean. Compute the Z-score using the formula: \(Z = \frac{(X - \mu)}{\sigma/\sqrt{n}}\). Substitute the values: \(Z = \frac{(75.75 - 75.5)}{3.5/\sqrt{6}}\). Calculate: \(Z \approx 0.175\).
05

Find the Probability Using Standard Normal Distribution

Using a Z-table, or a calculator with a normal distribution function, find the probability that corresponds to a Z-score of 0.175. This will give the probability of the sample mean being less than 75.75 psi.
06

Calculate the Probability of Exceeding the Given Value

Since we need the probability that the mean exceeds 75.75 psi, subtract the value found in Step 5 from 1. Assuming the cumulative probability from Step 5 is approximately 0.569, we find the probability of exceeding 75.75 psi is \(1 - 0.569 = 0.431\).

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Key Concepts

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

Normal Distribution
The concept of a normal distribution is fundamental in statistics, representing a data set that forms a symmetrical bell-shaped curve when graphed.
Each point along this curve corresponds to a probability density at which the data can occur.
  • The peak of the curve, its highest point, represents the mean of the data set.
  • The further you move away from the mean, the lower the probability density becomes, as indicated by the sides tapering off symmetrically.
  • Mean (渭): This is the average of the data set and is located at the center of the distribution.
  • Standard deviation (蟽): A measure of the data's variability, indicating how spread out the data points are around the mean.
Given that the tensile strength of the fiber specimens is normally distributed, we can use these properties to make probabilistic predictions about sample means.
Sampling Distribution
Sampling distribution is a probability distribution of a statistic, like a sample mean, based on random samples.
It tells us how the statistic behaves across different samples, providing a broader understanding of variability.
  • When studying a sampling distribution, the central limit theorem informs us that, regardless of the population distribution, the sampling distribution of the sample mean tends towards a normal distribution as the sample size increases.
  • For the exercise, the sample mean of six fiber specimens creates a sampling distribution, allowing us to make inferences about the general tensile strength.
  • The mean of the sampling distribution remains the same as the population mean (渭), in this case, 75.5 psi.
  • However, the standard deviation鈥攁 measure called the standard error鈥攁djusts according to the sample size.
Z-score
The Z-score is a measure that describes a value's position relative to the mean of a distribution.
Specifically, it tells us how many standard deviations away a particular value is from the mean.
  • In the context of sampling distributions, the Z-score formula is adjusted to account for the standard error rather than the population standard deviation.
  • Compute the Z-score using the formula: \[Z = \frac{(X - \mu)}{\sigma / \sqrt{n}}\]where \(X\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
  • For this problem, we find the Z-score of 0.175, indicating that the sample mean of 75.75 psi is 0.175 standard deviations above the mean of 75.5 psi.
The Z-score's calculated value then permits us to convert these standard deviations into probabilities using a Z-table or statistical software.
Standard Error
Standard error reflects the variability of a sample statistic, such as a sample mean, and is used within the context of a sampling distribution.
  • To find the standard error, divide the population's standard deviation by the square root of the sample size: \[SE = \frac{\sigma}{\sqrt{n}}\]
  • In the example, the standard deviation of 3.5 psi is divided by the square root of the sample size, yielding a standard error of approximately 1.428 psi.
  • This smaller standard error compared to the population standard deviation indicates less variability in the sample mean, facilitating more precise probability estimates.
  • The smaller the standard error, the closer the sample mean is likely to be to the population mean.
Standard error is crucial for constructing confidence intervals and understanding the reliability of sample statistics.

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Most popular questions from this chapter

\( \bar{X}_{1}\) and \(S_{1}^{2}\) are the sample mean and sample variance from a population with mean \(\mu_{1}\) and variance \(\sigma_{1}^{2}\). Similarly, \(\bar{X}_{2}\) and \(S_{2}^{2}\) are the sample mean and sample variance from a second independent population with mean \(\mu_{2}\) and variance \(\sigma_{2}^{2}\). The sample sizes are \(n_{1}\) and \(n_{2},\) respectively. (a) Show that \(\bar{X}_{1}-\bar{X}_{2}\) is an unbiased estimator of \(\mu_{1}-\mu_{2}\). (b) Find the standard error of \(\bar{X}_{1}-\bar{X}_{2}\). How could you estimate the standard error? (c) Suppose that both populations have the same variance; that is, \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}\). Show that $$ S_{p}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

The compressive strength of concrete is normally distributed with \(\mu=2500\) psi and \(\sigma=50\) psi. Find the probability that a random sample of \(n=5\) specimens will have a sample mean diameter that falls in the interval from 2499 psi to 2510 psi.

A computer software package calculated some numerical summaries of a sample of data. The results are displayed here:\begin{tabular}{cccccc} Variable & \(N\) & Mean & SE Mean & StDev & Variance \\ \hline\(x\) & 20 & 50.184 & \(?\) & 1.816 & \(?\) \end{tabular} (a) Fill in the missing quantities. (b) What is the estimate of the mean of the population from which this sample was drawn?

Like hurricanes and earthquakes, geomagnetic storms are natural hazards with possible severe impact on the Earth. Severe storms can cause communication and utility breakdowns, leading to possible blackouts. The National Oceanic and Atmospheric Administration beams electron and proton flux data in various energy ranges to various stations on the Earth to help forecast possible disturbances. The following are 25 readings of proton flux in the \(47-68\) kEV range (units are in \(\mathrm{p} /(\mathrm{cm} 2\) -sec-ster\(\mathrm{MeV}\) ) on the evening of December 28,2011: \(\begin{array}{llllllll}2310 & 2320 & 2010 & 10800 & 2190 & 3360 & 5640 & 2540 & 3360\end{array}\) \(\begin{array}{llllllll}11800 & 2010 & 3430 & 10600 & 7370 & 2160 & 3200 & 2020 & 2850\end{array}\) \(\begin{array}{lllllll}3500 & 10200 & 8550 & 9500 & 2260 & 7730 & 2250\end{array}\) (a) Find a point estimate of the mean proton flux in this time period. (b) Find a point estimate of the standard deviation of the proton flux in this time period. (c) Find an estimate of the standard error of the estimate in part (a). (d) Find a point estimate for the median proton flux in this time period. (e) Find a point estimate for the proportion of readings that are less than \(5000 \mathrm{p} /(\mathrm{cm} 2\) -sec-ster-MeV \()\).

Left right arrow A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval \(47 \leq X \leq 53\). Is the assumption of normality important? Why?
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