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91Ó°ÊÓ

In the United States, the mean age of men when they marry for the first time follows the normal distribution with a mean of 29 years. The standard deviation of the distribution is 2.5 years. For a random sample of 60 men, what is the likelihood that the age when they were first married is less than 29.3 years?

Short Answer

Expert verified
The probability is approximately 0.8238.

Step by step solution

01

Understand the Problem

We are given a mean (\( \mu \)) of age at first marriage as 29 years and a standard deviation (\( \sigma \)) of 2.5 years. We need to find the probability that the sample mean age of first marriage for 60 men is less than 29.3 years.
02

Determine the Z-score

To find the probability, we will use the Z-score formula for a sample mean: \[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]where \( \bar{x} = 29.3 \), \( \mu = 29 \), \( \sigma = 2.5 \), and \( n = 60 \).
03

Calculate the Standard Error

Calculate the standard error by dividing the standard deviation by the square root of the sample size: \[ \text{Standard Error} = \frac{2.5}{\sqrt{60}} \approx 0.3228 \]
04

Compute the Z-score

Substitute the values into the Z-score formula: \[ Z = \frac{29.3 - 29}{0.3228} \approx 0.93 \]
05

Find the Probability from the Z-score Table

Use a Z-score table to find the probability corresponding to a Z-score of 0.93. On most Z-tables, a Z-score of 0.93 approximates a probability of 0.8238.

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Key Concepts

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

Z-score
The Z-score is a critical concept in statistics, especially when dealing with normal distributions. A Z-score measures how many standard deviations an element is from the mean.
In simpler terms, it tells us where a particular data point is located within a distribution.
When calculating a Z-score, the formula is: \[Z = \frac{X - \mu}{\sigma}\]where:
  • \(X\) is the data point or sample mean we're interested in
  • \(\mu\) is the population mean
  • \(\sigma\) is the standard deviation
For a sample mean, you consider the sample size (n) as well, and the formula becomes:\[Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\]This adjusted formula accounts for variability in the sample mean.
The Z-score helps us find probabilities and make inferences about populations in standard ways.
Standard Error
The Standard Error (SE) measures the variability of a sample mean. It tells us how much the sample mean would vary if you took multiple samples from the same population.
When considering a sample mean, a smaller standard error indicates that our sample mean is close to the population mean.
The formula for SE when you have the standard deviation of the population is:\[SE = \frac{\sigma}{\sqrt{n}}\]where:
  • \(\sigma\) is the standard deviation of the population
  • \(n\) is the sample size
The standard error helps in determining how well the sample mean estimates the population mean.
In the given problem, it plays a crucial part in finding the Z-score and thus the probability of our event.
Probability
Probability is a way to describe uncertainty. It measures the likelihood of a particular outcome occurring.
In the context of normal distribution, we often use probability to understand where a sample mean lies in relation to the population mean.
Once we have the Z-score, we can use a Z-table to find the probability.
A Z-table provides the percentage of data below a particular Z-score in a standard normal distribution.
In our exercise, a Z-score of 0.93 translates to a probability of 0.8238, indicating that approximately 82.38% of data points fall below the sample mean age we are analyzing. Probability helps us interpret results and make decisions based on statistical evidence.
Sample Mean
The Sample Mean is the average of a set of data points collected from a portion of the entire population.
It provides an estimate of the population mean. In statistics, sample means are crucial for making inferences about populations. The formula for the sample mean is straightforward: \[\bar{x} = \frac{\sum x_i}{n}\]where:
  • \(\bar{x}\) is the sample mean
  • \(\sum x_i\) is the sum of all data points in the sample
  • \(n\) is the number of data points in the sample
Understanding the sample mean is essential when predicting or estimating population characteristics, as shown in our problem where we used it to find the probability that the mean age is less than a certain value.
Sample means allow statisticians to perform analyses when it's not feasible to study the entire population.

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Most popular questions from this chapter

A population consists of the following five values: \(2,2,4,4,\) and 8 . a. List all samples of size 2 , and compute the mean of each sample. b. Compute the mean of the distribution of sample means and the population mean. Compare the two values. c. Compare the dispersion in the population with that of the sample means.

Suppose your statistics instructor gave six examinations during the semester. You received the following exam scores (percent correct): \(79,64,84,82,92,\) and 77 . The instructor decided to randomly select two exam scores, compute their mean, and use this score to determine your final course grade. a. Compute the population mean. b. How many different samples of two test grades are possible? c. List all possible samples of size 2 and compute the mean of each. d. Compute the mean of the sample means and compare it to the population mean. e. If you were a student, would you like this arrangement? Would the result be different from dropping the lowest score? Write a brief report.

Nike's annual report says that the average American buys 6.5 pairs of sports shoes per year. Suppose a sample of 81 customers is surveyed and the population standard deviation of sports shoes purchased per year is 2.1 . a. What is the standard error of the mean in this experiment? b. What is the probability that the sample mean is between 6 and 7 pairs of sports shoes? c. What is the probability that the difference between the sample mean and the population mean is less than 0.25 pair? d. What is the likelihood the sample mean is greater than 7 pairs?

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is \(\$ 165,000 .\) This distribution follows the normal distribution with a standard deviation of \(\$ 40,000 .\) a. If we select a random sample of 50 households, what is the standard error of the mean? b. What is the expected shape of the distribution of the sample mean? c. What is the likelihood of selecting a sample with a mean of at least \(\$ 167,000 ?\) d. What is the likelihood of selecting a sample with a mean of more than \(\$ 155,000 ?\) e. Find the likelihood of selecting a sample with a mean of more than \(\$ 155,000\) but less than \(\$ 167,000\).

An economist uses the price of a gallon of milk as a measure of inflation. She finds that the average price is \(\$ 3.82\) per gallon and the population standard deviation is \(\$ 0.33 .\) You decide to sample 40 convenience stores, collect their prices for a gallon of milk, and compute the mean price for the sample. a. What is the standard error of the mean in this experiment? b. What is the probability that the sample mean is between \(\$ 3.78\) and \(\$ 3.86 ?\) c. What is the probability that the difference between the sample mean and the population mean is less than \(\$ 0.01 ?\) d. What is the likelihood the sample mean is greater than \(\$ 3.92 ?\)

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