two different sections of electricians took their state license test. the first section consisted of 75 applicants having an average score of 262 and a sample standard deviation of 35. they were journeyman, experienced electricians who were renewing their licenses. the second section consisted of 34 applicants with an average score of 233 and a sample standard deviation of 27. they were apprenticed electricians who were just beginning their careers.

Question

two different sections of electricians took their state license test. the first section consisted of 75 applicants having an average score of 262 and a sample standard deviation of 35. they were journeyman, experienced electricians who were renewing their licenses. the second section consisted of 34 applicants with an average score of 233 and a sample standard deviation of 27. they were apprenticed electricians who were just beginning their careers.

the distributions of scores during previous years have shown that licensing exam scores are normally distributed. Based on this information, which of the following best estimates the number of applicants that passed the test, if an applicant must score 200 or greater to pass?

a) 102
b) 84
c) 159
d) 148
e)17

Answer 1

To estimate the number of applicants that passed the test, we need to calculate the z-scores for both sections and find the proportion of applicants who scored 200 or greater.

First, let's calculate the z-score for the first section using the formula: (score - mean) / standard deviation.

For the first section:
Z = (200 - 262) / 35
Z = -1.771

Next, we need to find the proportion of applicants with a z-score greater than or equal to -1.771. We can look up this proportion in a standard normal distribution table or use a statistics calculator.

Using a standard normal distribution table, the proportion of applicants with a z-score of -1.771 or greater is approximately 0.9612.

Now, let's calculate the z-score for the second section using the same formula: (score - mean) / standard deviation.

For the second section:
Z = (200 - 233) / 27
Z = -1.222

Again, we need to find the proportion of applicants with a z-score greater than or equal to -1.222. Using a standard normal distribution table, the proportion is approximately 0.8888.

To estimate the number of applicants that passed the test, we need to add up the proportions for both sections and multiply by the total number of applicants in each section.

For the first section: 0.9612 * 75 = 72.09 (rounded to 72)

For the second section: 0.8888 * 34 = 30.20 (rounded to 30)

Now, summing up both sections: 72 + 30 = 102

Therefore, the best estimate for the number of applicants who passed the test is 102.

The correct answer is (a) 102.