Question

A variable is normally distributed with a mean of 120 and a standard deviation of 5 . Four scores are randomly sampled. What is the probability that the mean of the four scores is above \(127 ?\)

Short answer

0.26%

Step-by-step solution

Identify the Question Requirements

We need to find the probability that the mean of four randomly sampled scores is greater than 127. We have a normally distributed variable with a mean (\( \mu \) = 120) and a standard deviation (\( \sigma \) = 5).

Calculate the Standard Error of the Mean

The standard error of the mean (SEM) is calculated by dividing the standard deviation by the square root of the sample size (\(n = 4\)). Therefore: \( SEM = \frac{5}{\sqrt{4}} = 2.5 \).

Convert the Score to a Z-score

To find the probability, convert the sample mean to a z-score using the formula: \( Z = \frac{\bar{X} - \mu}{SEM} \). Here, \( \bar{X} = 127 \), \( \mu = 120 \), and \( SEM = 2.5 \). So, \( Z = \frac{127 - 120}{2.5} = 2.8 \).

Use Z-table to Find Probability

Use a standard normal distribution table (z-table) to find the probability corresponding to \( Z = 2.8 \). The probability in the table for \( Z = 2.8 \) is approximately 0.9974. We are interested in the probability that the mean is above 127, so we find: \( 1 - 0.9974 = 0.0026 \).

Conclude the Result

The probability that the mean of the four scores is above 127 is 0.0026 or 0.26%.

Key concepts

Normal Distribution

In probability theory, the normal distribution is a key concept often used to model real-world phenomena. You might hear it referred to as a "bell curve" because of its symmetrical, bell-shaped appearance when graphed. Here are some important features of a normal distribution:

• It is defined by two parameters: the mean (average) and the standard deviation (which measures spread or variability).
• The mean determines where the center of the curve sits on the horizontal axis.
• The standard deviation determines the width of the curve. A smaller standard deviation makes the curve narrower, while a larger one widens it.

Normal distribution is useful in the exercise provided because the problem involves a variable that is normally distributed with a mean of 120 and a standard deviation of 5. This allows us to leverage the properties of the normal distribution to calculate probabilities for sample means.

Z-score

The z-score is a statistical measurement that describes a value's relation to the mean of a group of values. When dealing with a normal distribution, it helps us understand how far away a particular data point is from the mean. Here's the essence of how z-scores work:

• A z-score tells you how many standard deviations a value is from the mean.
• Values above the mean have positive z-scores, while those below have negative z-scores.

To find the z-score for this exercise, you use the formula: \[ Z = \frac{\bar{X} - \mu}{SEM} \]where \( \bar{X} \) is the sample mean (127 in this case), \( \mu \) is the population mean (120), and SEM is the standard error of the mean. Calculating this gives us a z-score of 2.8. This tells us that the sample mean is 2.8 standard errors above the mean.

Standard Error of the Mean

The standard error of the mean (SEM) is an important concept in probability and statistics, especially when dealing with sample data. It provides a measure of how much variability can be expected in a sample mean if you were to draw multiple samples from the same population.The SEM is calculated by dividing the standard deviation by the square root of the sample size (\( n \)). In mathematical terms:\[ SEM = \frac{\sigma}{\sqrt{n}} \]where \( \sigma \) is the standard deviation of the population. For the given exercise, the standard deviation is 5, and the sample size is 4. Therefore, the SEM is calculated as \( 2.5 \). This smaller value compared to the standard deviation indicates less variability between sample means and the population mean due to the averaging effect of the sample size.