Normal Distribution and Sampling Distributions: Probability of Sample Mean

Normal Distribution and Continuous Random Variables

Sampling Distribution of the Mean

The sampling distribution of the mean describes the distribution of sample means for a given sample size, drawn from a population with a known mean and standard deviation. When the population is normally distributed, the sample mean is also normally distributed.

• Population Mean (μ): The average value in the population. In this problem, μ = 98.20°F.

• Population Standard Deviation (σ): The measure of spread in the population. Here, σ = 0.62°F.

• Sample Size (n): The number of observations in the sample. Here, n = 19.

• Sampling Distribution Mean: Equal to the population mean, μ.

• Sampling Distribution Standard Deviation (Standard Error): $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

Calculating the Probability for the Sample Mean

To find the probability that the mean body temperature of 19 randomly selected people is less than 98.50°F, we use the properties of the normal distribution and the sampling distribution of the mean.

• Step 1: Calculate the Standard Error $\sigma_{\bar{x}} = \frac{0.62}{\sqrt{19}}$ $\sigma_{\bar{x}} \approx 0.142$

• Step 2: Find the Z-score for the sample mean $Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$ $Z = \frac{98.50 - 98.20}{0.142}$ $Z \approx 2.11$

• Step 3: Use the Z-score to find the probability The probability that the sample mean is less than 98.50°F is the area to the left of Z = 2.11 in the standard normal distribution. From standard normal tables: $P(Z < 2.11) \approx 0.9826$

Example

Problem: Human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.62°F. If 19 people are randomly selected, what is the probability that their mean body temperature is less than 98.50°F?

• Solution:

  1. Calculate the standard error: $\sigma_{\bar{x}} = \frac{0.62}{\sqrt{19}} \approx 0.142$

  2. Compute the Z-score: $Z = \frac{98.50 - 98.20}{0.142} \approx 2.11$

  3. Find the probability: $P(Z < 2.11) \approx 0.9826$

• Interpretation: There is approximately a 98.26% chance that the mean body temperature of a random sample of 19 people will be less than 98.50°F.

Key Formulas

• Standard Error of the Mean: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

• Z-score for Sample Mean: $Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$

Additional info:

• The problem is a classic application of the Central Limit Theorem, which states that the sampling distribution of the sample mean approaches normality as the sample size increases, even if the population distribution is not normal. In this case, the population is already normal.

• Standard normal tables or calculators are used to find the probability associated with a given Z-score.