A variable is normally distributed with a mean of 100 and a standard deviation of 10. What percentage of observations of the variable lie between 106 and 110?

To answer this question, we need to use the properties of the standard normal distribution (Z-distribution).

Step 1: Convert the raw scores to Z-scores. The Z-score is a measure of how many standard deviations an element is from the mean.

The formula for calculating a Z-score is: Z = (X - μ) / σ

Where:

• X is the raw score
• μ is the mean
• σ is the standard deviation

For X = 106, the Z-score is: Z = (106 - 100) / 10 = 0.6 For X = 110, the Z-score is: Z = (110 - 100) / 10 = 1

Step 2: Use the standard normal distribution table (Z-table) to find the percentage of observations that lie between these two Z-scores.

The Z-table tells us that:

• The percentage of observations that lie below a Z-score of 0.6 is approximately 72.57%.
• The percentage of observations that lie below a Z-score of 1 is approximately 84.13%.

Step 3: Subtract the two percentages to find the percentage of observations that lie between the two Z-scores.

So, the percentage of observations that lie between a Z-score of 0.6 and 1 is approximately 84.13% - 72.57% = 11.56%.

Therefore, approximately 11.56% of observations of the variable lie between 106 and 110.