Question 213459: A random sample of 100 students at Southern University was collected. The mean age was found to be 24.2 years, with a standard deviation of 3.4 years.
(a) What percentage of students are over 27.6 years old? 4%
(b) What percentage of the students are between 20.8 and 24.2 years old? 9%
(c) If a student is selected at random from Southern University, what is the probability that he or she will be over 20.8 years old? 5%
Ignore the 4 percent, 9 percent and the 5 percent that was what I think the answer is.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A random sample of 100 students at Southern University was collected. The mean age was found to be 24.2 years, with a standard deviation of 3.4 years.
(a) What percentage of students are over 27.6 years old?
Find the z-score of 27.6.
z(27.6) = (27.6-24.2)/3.4 = 1
Then P(x > 27.6) = P(z > 1) = 0.1587
==============================================
(b) What percentage of the students are between 20.8 and 24.2 y0.34ears old?
Find the z-scores of the two numbers:
z(24.2) = 0
z(20.8) = (20.8-24.2) = -3.4=2.4 = -1
P(20.8 < x < 24.2) = P(-1 < z < 0) = 0.3413
-----------------------------------------------------------
(c) If a student is selected at random from Southern University, what is the probability that he or she will be over 20.8 years old?
z(20.8 = -1
P(x > 20.8) = P(z > -1) = 0.8413
=========================================
Cheers,
stan H.
|
|
|
