Complete the following question using Excel. A large company gives each of its e
ID: 3365903 • Letter: C
Question
Complete the following question using Excel.
A large company gives each of its employees an aptitude test. The scores on the test are normally distributed with a mean of 77 and a standard deviation of 14. A sample of 24 employees is taken.
Is the distribution of the sample means normal? Enter "yes" or "no" in blank #1.
In blank #2 enter the probability that the test score for a randomly selected employee is between 75 and 79. (Round your answer to 4 decimal places and enter your answer as 0.1234 not .1234.)
In blank #3 enter the probability that the average test score in the sample is less than 78.93. (Round your answer to 4 decimal places and enter your answer as 0.1234 not .1234.)
In blank #4 enter the probability that the average test score in the sample is at least 80.25. (Round your answer to 4 decimal places and enter your answer as 0.1234 not .1234.)
In blank #5 enter the probability that the average test score in the sample is between 73.65 and 81.65. (Round your answer to 4 decimal places and enter your answer as 0.1234 not .1234.)
Explanation / Answer
Is the distribution of the sample means normal? Enter "yes" or "no" in blank #1.
Ans = yes
In blank #2 enter the probability that the test score for a randomly selected employee is between 75 and 79
Z = (X - 77)/14
P(75 <X< 79)
= P(-0.1429 < Z< 0.1429)
= 0.11359 = 0.1136
In blank #3 enter the probability that the average test score in the sample is less than 78.93.
sd(Xbar) = 14/sqrt(24) = 2.85773
P(Xbar < 78.93)= P(Z < 0.675359) = 0.750276 = 0.7503
In blank #4 enter the probability that the average test score in the sample is at least 80.25.
P(Xbar > 80.25)
= P(Z > 1.137263) = 0.127714 = 0.1277
In blank #5 enter the probability that the average test score in the sample is between 73.65 and 81.65.
P(73.65 <Xbar< 81.65)
= P(-1.17226 <Z< 1.627161)
= 0.8276