Question

In a normal distribution of test scores with a mean of 100 and a standard deviation of 7, what is (are) the score(s): a. where the bottom 5% of the cases are located? b. where the top 52% of the cases.

Answer 1

The score for the bottom 5% of cases in a normal distribution with mean 100 and standard deviation 7 is approximately 88.5. For the top 52% of cases, the approximate score is 99.65, using z-scores and the properties of normal distribution.

Step-by-step explanation:

The student is asking about locating specific scores in a normal distribution with a given mean and standard deviation. To find these scores, we need to use the concept of z-scores and the properties of the normal distribution.

Part (a): Bottom 5% of cases

To find the score where the bottom 5% of cases are located, we look up the z-score that corresponds to the 5th percentile in a standard normal distribution table. The z-score associated with the bottom 5% is approximately -1.645. We then use the formula:

Score = Mean + (Z-score × Standard Deviation)

Substituting the given mean of 100 and standard deviation of 7, we get:

Score = 100 + (-1.645 × 7) = 100 - 11.515 = 88.485

The score where the bottom 5% of cases are located is approximately 88.5.

Part (b): Top 52% of cases

To find the score where the top 52% of cases are located, we need to find the z-score associated with the bottom 48% (because the normal distribution is symmetric). The z-score for the 48th percentile is roughly -0.05. Again, using the same formula:

Score = 100 + (-0.05 × 7) = 100 - 0.35 = 99.65

The score corresponding to the top 52% of cases is approximately 99.65.