Question

A​ bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. Find ​, the score which separates the lower​ 60% from the top​ 40%. Round to one decimal place.

Answer 1

The score separating the lower 60% from the top 40% is found using the normal distribution with the given mean and standard deviation. The 60th percentile corresponds to a z-score of 0.25, which translates to a score of 212.5 when applying the normal distribution formula.

Step-by-step explanation:

The question is asking to find the score that separates the lower 60% from the top 40% of the bank's applicants, where the loan officer rates have a normal distribution with a mean (μ) of 200 and a standard deviation (σ) of 50. To find this value, known as the 60th percentile, we use a z-score table or normal distribution calculator.

To begin, we must convert the percentile to a z-score. From a standard normal distribution table, a lower 60% corresponds to a z-score of approximately 0.25. We then use the formula:

score = mean + (z-score * standard deviation)

Therefore:

score = 200 + (0.25 * 50) = 200 + 12.5 = 212.5

So, the score that separates the lower 60% from the top 40% is approximately 212.5, rounded to one decimal place.