Question

At a company, the data set containing the ages of applicants for a particular job was 30 years old, and the standard deviation of the data set was 3.5 years. What is the closest percent of applicants that were 21 years old or younger?

Answer 1

Using the normal distribution model, we calculate that approximately 0.5% of applicants are 21 years old or younger. The calculation involves finding the Z-score for age 21 and then looking up the corresponding cumulative probability.

Step-by-step explanation:

To find the closest percent of applicants that were 21 years old or younger, given the mean age is 30 years old with a standard deviation of 3.5 years, we can use the normal distribution model. To do this, we first need to calculate how many standard deviations away 21 is from the mean. This is done by taking the difference between the target age and the mean, then dividing by the standard deviation: (21 - 30) / 3.5 = -9 / 3.5 = -2.57. Next, we can use the Z-table to find the area to the left of -2.57. This area represents the proportion (and thus the percent) of applicants who are 21 years old or younger.

From standard normal distribution tables, a Z-score of -2.57 has an associated cumulative probability of approximately 0.005. This means that only about 0.5% of applicants would be expected to be 21 years old or younger.