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Chapter 7: Problem 58

A random sample of size \(n=16\) is taken from a normal population with \(\mu=40\) and \(\sigma^{2}=5 .\) Find the probability that the sample mean is less than or equal to \(37 .\)

Short Answer

Expert verified
The probability that the sample mean is less than or equal to 37 is effectively 0.

Step by step solution

01

Identify the known values

We have a random sample of size \( n = 16 \) drawn from a normal population with a mean \( \mu = 40 \) and variance \( \sigma^2 = 5 \). We need to find the probability that the sample mean \( \bar{x} \) is less than or equal to 37.
02

Determine the sampling distribution of the sample mean

Since the sample is from a normal population, the sample mean \( \bar{x} \) will also be normally distributed with mean \( \mu_{\bar{x}} = \mu = 40 \) and standard deviation \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{5}}{4} \), where \( \sigma = \sqrt{5} \) and \( n = 16 \).
03

Calculate standard deviation of the sample mean

The standard deviation for the sample mean \( \bar{x} \) is \( \sigma_{\bar{x}} = \frac{\sqrt{5}}{4} = \frac{\sqrt{5}}{4} \approx 0.559 \).
04

Find the Z-score for the sample mean

We want to find \( P(\bar{x} \leq 37) \). Calculate the Z-score for \( \bar{x} = 37 \):\[ Z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{37 - 40}{0.559} \approx -5.368 \]
05

Use Z-score to find the probability

Using the Z-score \( Z \approx -5.368 \), check the standard normal distribution to find \( P(Z \leq -5.368) \). For practical purposes, this value is extremely small and is effectively \(0\), since the Z-score table can't reach that far in either direction.

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Key Concepts

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

Normal Distribution
A normal distribution is a core concept in statistics that represents a symmetric, bell-shaped curve. The curve is defined by two parameters: the mean (\(\mu\)) and the variance (\(\sigma^2\)). The mean determines the center of the distribution, and the variance indicates how spread out the values are around the mean.

In a normal distribution, most of the data points tend to cluster around the mean, while fewer are found in the tails. About 68% of the data lies within one standard deviation (\(\sigma\)) of the mean, 95% within two, and 99.7% within three.

This property is known as the empirical rule or the 68-95-99.7 rule. Given the prevalence of normal distribution in real-world data, understanding it is crucial as it's the foundation for many statistical analyses.

In this exercise, the sample is drawn from a normal population with a mean of 40 and a variance of 5, setting the background for further calculations.
Z-score
The Z-score helps us understand how far a particular score or data point is from the mean, with reference to the standard deviation. It's a way of standardizing scores across different datasets or distributions.

Given by the formula:\[Z = \frac{X - \mu}{\sigma}\]where:
  • \(X\) is the value of interest (in our exercise, the sample mean),
  • \(\mu\) is the mean of the distribution,
  • \(\sigma\) is the standard deviation.


For example, a Z-score of 1 means the value is one standard deviation above the mean. A Z-score of -5.368 indicates the value is significantly below the mean, as seen in our exercise. Using the Z-score, we convert the value into a standardized form, which helps in determining probabilities using the standard normal distribution table.
Sample Mean
The sample mean, often denoted as \(\bar{x}\), is an average of all the data points in a sample. It's used to estimate the population mean, especially when the population is too large to measure entirely.

Importantly, if the sample is taken from a normal population, the sample mean itself follows a normal distribution. This is the foundation of many inferential statistics techniques.

The mean of the sample mean's distribution is equal to the population mean (\(\mu\)), and its standard deviation, often called the standard error, is given by:\[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\]where \(\sigma\) is the population standard deviation and \(n\) is the sample size.

In this exercise, with a sample size of 16, we can calculate the standard deviation of the sample mean, which is crucial for finding the Z-score.
Probability Calculation
Probability calculation involves determining the likelihood that a particular event will occur. In statistics, especially with normal distributions, it's often about finding the probability associated with a specified range of values for a normally distributed random variable.

In this exercise, we need the probability that the sample mean is less than or equal to 37. Using the calculated Z-score of -5.368, we then refer to the standard normal distribution table.

Such tables, or software tools, provide the probability of a value falling below a given Z-score. Since -5.368 is far in the tail, the probability is nearly zero, indicating it's highly unlikely for the sample mean to be this low if the population mean is truly 40. Understanding this step is crucial in fields that rely heavily on data-driven decision-making, ensuring accurate risk assessment and predictions.

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Most popular questions from this chapter

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 \(47 \leq \bar{X} \leq 53 .\) Is the assumption of normality important? Why?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be uniformly distributed on the interval 0 to \(a\). Recall that the maximum likelihood estimator of \(a\) is \(\hat{a}=\max \left(X_{i}\right)\) (a) Argue intuitively why \(\hat{a}\) cannot be an unbiased estimator for \(a\) (b) Suppose that \(E(\hat{a})=n a /(n+1)\). Is it reasonable that \(\hat{a}\) consistently underestimates \(a\) ? Show that the bias in the estimator approaches zero as \(n\) gets large. (c) Propose an unbiased estimator for \(a\). (d) Let \(Y=\max \left(X_{i}\right)\). Use the fact that \(Y \leq y\) if and only if each \(X_{i} \leq y\) to derive the cumulative distribution function of \(Y\). Then show that the probability density function of \(Y\) is $$f(y)=\left\\{\begin{array}{cl}\frac{n y^{n-1}}{a^{n}}, & 0 \leq y \leq a \\\0, & \text { otherwise }\end{array}\right.$$ Use this result to show that the maximum likelihood estimator for \(a\) is biased. (e) We have two unbiased estimators for \(a:\) the moment estimator \(\hat{a}_{1}=2 \bar{X}\) and \(\hat{a}_{2}=[(n+1) / n] \max \left(X_{i}\right)\) where \(\max \left(X_{i}\right)\) is the largest observation in a random sample of size \(n\). It can be shown that \(V\left(\hat{a}_{1}\right)=a^{2} /(3 n)\) and that \(V\left(\hat{a}_{2}\right)=a^{2} /[n(n+2)] .\) Show that if \(n>1, \hat{a}_{2}\) is a better estimator than \(\hat{a}\). In what sense is it a better estimator of \(a\) ?

When the sample standard deviation is based on a random sample of size \(n\) from a normal population, it can be shown that \(S\) is a biased estimator for \(\sigma\). Specifically, $$E(S)=\sigma \sqrt{2 /(n-1)} \cdot \Gamma(n / 2) / \Gamma[(n-1) / 2]$$ (a) Use this result to obtain an unbiased estimator for \(\sigma\) of the form \(c_{n} S\), when the constant \(c_{n}\) depends on the sample size \(n\). (b) Find the value of \(c_{n}\) for \(n=10\) and \(n=25\). Generally, how well does \(S\) perform as an estimator of \(\sigma\) for large \(n\) with respect to bias?

Consider the shifted exponential distribution $$f(x)=\lambda e^{-\lambda(x-\theta)}, \quad x \geq \theta$$ When \(\theta=0,\) this density reduces to the usual exponential distribution. When \(\theta>0,\) there is only positive probability to the right of \(\theta\) (a) Find the maximum likelihood estimator of \(\lambda\) and \(\theta,\) based on a random sample of size \(n\). (b) Describe a practical situation in which one would suspect that the shifted exponential distribution is a plausible model.

A collection of \(n\) randomly selected parts is measured twice by an operator using a gauge. Let \(X_{i}\) and \(Y_{i}\) denote the measured values for the \(i\) th part. Assume that these two random variables are independent and normally distributed and that both have true mean \(\mu_{i}\) and variance \(\sigma^{2}\) (a) Show that the maximum likelihood estimator of \(\sigma^{2}\) is \(\hat{\sigma}^{2}=(1 / 4 n) \sum_{i=1}^{n}\left(X_{i}-Y_{i}\right)^{2}\) (b) Show that \(\hat{\sigma}^{2}\) is a biased estimator for \(\hat{\sigma}^{2}\). What happens to the bias as \(n\) becomes large? (c) Find an unbiased estimator for \(\sigma^{2}\).
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