Question

A distribution of values is normal with a mean of 100 and a standard deviation of 6. Find the interval containing the middle-most 60 s of scores: Enter your answer using interval notation. Example: [2,1,5,6) Your numbers should be accurate to 1 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

The interval containing the middle-most 60% of scores for a normal distribution with mean 100 and standard deviation 6 is approximately [95.0, 105.0] using z-scores and the standard normal distribution table.

Step-by-step explanation:

The student is asked to find the interval containing the middle-most 60% of scores for a normal distribution with a mean of 100 and a standard deviation of 6. To find this interval, we need to use the z-scores associated with the middle 60%. Since we're not given the exact z-scores to use for 60%, we need to look at a standard normal distribution table or use a calculator to determine the z-scores that correspond to the 20th percentile (bottom 20%) and the 80th percentile (top 20%), thus encompassing the middle 60%.

The z-score for the bottom 20% is approximately -0.84 and for the top 20% is approximately +0.84. We can then convert these z-scores back to the original scale using the formula:

x = μ + (z)(σ)

For the lower limit: x = 100 + (-0.84)(6) = 94.96 (approximately 95)

For the upper limit: x = 100 + (0.84)(6) = 105.04 (approximately 105)

The interval containing the middle-most 60% of scores is approximately [95.0, 105.0] in interval notation.