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Chapter 5: Problem 59

Find the indicated probabilities and interpret the results. The mean annual salary for intermediate level life insurance underwriters is about \(\$ 61,000 .\) A random sample of 45 intermediate level life insurance underwriters is selected. What is the probability that the mean annual salary of the sample is (a) less than \(\$ 60,000\) and (b) more than \(\$ 63,000 ?\) Assume \(\sigma=\$ 11,000\). (Adapted from Salary.com)

Short Answer

Expert verified
P(mean < 60,000) 鈮� 27.1%; P(mean > 63,000) 鈮� 11.1%.

Step by step solution

01

Identify the Known Values

The mean annual salary () for the population is \( \mu = 61,000 \). The standard deviation is \( \sigma = 11,000 \). The sample size is \( n = 45 \). We need to find the probabilities for the sample mean.
02

Calculate the Standard Error

The standard error (SE) of the sample mean is used to measure the dispersion of the sample means around the population mean. It is calculated as \( \text{SE} = \frac{\sigma}{\sqrt{n}} \). Quantitatively, \( \text{SE} = \frac{11,000}{\sqrt{45}} \approx 1,640.92 \).
03

Define the Z-Score Formulas

The Z-score helps in determining the number of standard deviations a data point is from the mean. The formula for the Z-score for the sample mean is given by \( Z = \frac{\bar{x} - \mu}{\text{SE}} \), where \( \bar{x} \) is the sample mean.
04

Calculate Z-Score for Part (a)

For \( \bar{x} = 60,000 \), the Z-score is \( Z = \frac{60,000 - 61,000}{1,640.92} \approx -0.61 \).
05

Find the Probability for Part (a)

Using the Z-score table or normal distribution calculator, find the probability (P) of \( Z < -0.61 \). \( P(Z < -0.61) \approx 0.2709 \).
06

Calculate Z-Score for Part (b)

For \( \bar{x} = 63,000 \), the Z-score is \( Z = \frac{63,000 - 61,000}{1,640.92} \approx 1.22 \).
07

Find the Probability for Part (b)

Using the Z-score table or normal distribution calculator, find the probability (P) of \( Z > 1.22 \). \( P(Z > 1.22) \approx 0.1112 \).
08

Interpret the Results

The probability that the mean annual salary of the sample is less than \\(60,000 is approximately 27.1%, and the probability that it is more than \\)63,000 is approximately 11.1%.

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Key Concepts

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

Standard Error
Standard Error (SE) is a critical concept in understanding how sample means relate to the population mean. It helps us measure how much variation or 'error' exists when estimating the population mean using a sample mean. In simpler terms, the Standard Error tells us how far the sample mean of the data is expected to be from the actual population mean.
  • It is especially useful when dealing with large sample sizes.
  • SE is smaller for larger samples, reflecting more reliable estimates of the population mean.
The formula for Standard Error is given by:\[\text{SE} = \frac{\sigma}{\sqrt{n}}\]where:
  • \(\sigma\) is the population standard deviation
  • \(n\) is the sample size

In our example, using \(\sigma = 11,000\) and \(n = 45\), the Standard Error is approximately 1,640.92. This value allows us to understand the range within which the sample means might fall.
Z-score
The Z-score is another fundamental concept in statistics used to understand the relationship of a data point to the mean of a data set. Specifically, it tells us how many Standard Errors we are away from the mean.
  • A Z-score can help determine the likelihood of a score falling above or below a particular value.
  • Positive Z-scores indicate the value is above the mean, while negative Z-scores show it's below the mean.
The formula for the Z-score when dealing with sample means is:\[Z = \frac{\bar{x} - \mu}{\text{SE}} \]where:
  • \(\bar{x}\) is the sample mean
  • \(\mu\) is the population mean
  • \(\text{SE}\) is the Standard Error

For instance, a Z-score of -0.61 for a sample mean of \(60,000\) suggests that this value is 0.61 Standard Errors below the population mean of \(61,000\).
Normal Distribution
Normal Distribution is a key pillar in statistics. It is often depicted as a symmetrical bell-shaped curve.
  • In real-world scenarios, many variables tend to follow this distribution.
  • The bell curve is essential for probability estimations and statistical inference.
Due to its properties, the mean, median, and mode of a normal distribution are equal.Standard deviation and mean define the curve's shape and center. In our context, it allows us to deduce probabilities for certain outcomes.By translating sample means into Z-scores, we can use the normal distribution to determine how often such sample means occur. For example, the probability that a sample mean is less than \(60,000\) can be easily found using the standard normal distribution, due to its predictable properties.
Sample Mean
Sample Mean refers to the average value in a subset (the sample) of a larger population.
  • It's used to make inferences about the population mean (\(\mu\)).
  • The sample mean is represented by \(\bar{x}\).
Using sample means can be efficient for understanding population characteristics without measuring the entire group. Although influenced by each sample's individual values, the larger the sample size, the closer the sample mean tends to be to the population mean.In statistical analysis, the sample mean forms the basis for calculating probabilities, as seen in our task. With a sample mean, we define Z-scores, helping us figure out the probabilities of those sample means occurring, based on the normal distribution.

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