Determining Mean and Standard Deviation in a Normal Distribution: A Practical Example

To determine the mean and standard deviation of a normal distribution, we often use the properties of the standard normal distribution and the concept of Z-scores. This article will walk you through a practical example where 31 of the items are under 45 and 8 items are over 64. By following the steps outlined below, you can solve for the mean and standard deviation.

Understanding the Problem

Given:

31 of the items are under 45: This means that the probability that a randomly selected item is less than 45 is 0.31. 8 of the items are over 64: This means that the probability that a randomly selected item is greater than 64 is 0.08, which is equivalent to the probability that a randomly selected item is less than 64 being 0.92.

Step 1: Find Z-Scores for the Given Probabilities

The Z-score is a measure of how many standard deviations an element is from the mean. We can find the Z-scores corresponding to the given probabilities using a Z-table or a calculator.

For P(X 45) 0.31: The Z-score corresponding to 0.31 is approximately -0.497.

For P(X 64) 0.92: The Z-score corresponding to 0.92 is approximately 1.405.

Step 2: Set Up the Equations

The Z-score formula is given by:

z (X - mu) / sigma

Therefore:

For X 45:

-0.497 (45 - mu) / sigma

For X 64:

1.405 (64 - mu) / sigma

Step 3: Solve the Equations

From equation 1:

45 - mu -0.497sigma mu 45 0.497sigma

From equation 2:

64 - mu 1.405sigma mu 64 - 1.405sigma

Equate the two expressions for mu:

45 0.497sigma 64 - 1.405sigma

1.902sigma 19 sigma approx 9.98 approx 10

Substitute sigma back to find mu:

mu 45 0.497 times 9.98 approx 45 4.96 approx 49.96 approx 50

Final Results

The mean mu is approximately 50, and the standard deviation sigma is approximately 10.

Therefore, the mean is about 50 and the standard deviation is about 10.

Conclusion: By understanding and applying the properties of the normal distribution, we can determine the mean and standard deviation even with limited information about the distribution.