Assume that at WMU the average engineering GPA is 2.9 with a standard deviation
ID: 2929011 • Letter: A
Question
Assume that at WMU the average engineering GPA is 2.9 with a standard deviation of 30. Assume that the population is normally distributed and that a sample of 50 students is selected. a. What is the probability that the sample average GPA is greater than 3.5? b. What is the probability that the sample average GPA is less than 3.0? c. What is the probability that the sample average GPA is between 2.8 and 3.2? d. There is a 90% probability that the sample average is above what GPA? 6. There is 75% probability that the sample average is below what GPA? In what scenarios, would you use the normal distribution and student t distribution to calculate a confidence interval for ? Explain fully 7. An average scanned image occupies 0.6 megabytes of memory with a standard deviation of 0.4 megabytes. a. If you just plan to install only one image on your web site, what is the probability that the size of the image is between 0.2 and 1 megabytes? b. If you plan to install 80 images on your web site, what is the probability that their total size is between 47 megabytes and 56 megabytes? (hint: think about central limit theorem, the total size of 80 images-80 * the average size of the 80 images) 8.Explanation / Answer
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 2.9
standard Deviation ( sd )= 0.3
sample size (n) = 50
a.
P(X > 3.5) = (3.5-2.9)/0.3/ Sqrt ( 50 )
= 0.6/0.042= 14.1421
= P ( Z >14.1421) From Standard Normal Table
= 0
b.
P(X < 3) = (3-2.9)/0.3/ Sqrt ( 50 )
= 0.1/0.0424= 2.357
= P ( Z <2.357) From Standard NOrmal Table
= 0.9908
c.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 2.8) = (2.8-2.9)/0.3/ Sqrt ( 50 )
= -0.1/0.0424
= -2.357
= P ( Z <-2.357) From Standard Normal Table
= 0.0092
P(X < 3.2) = (3.2-2.9)/0.3/ Sqrt ( 50 )
= 0.3/0.0424 = 7.0711
= P ( Z <7.0711) From Standard Normal Table
= 1
P(2.8 < X < 3.2) = 1-0.0092 = 0.9908
d.
90% of probability that sample average is above = P ( Z < x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is -1.2816
P( x-u/s.d < x - 2.9/0.0424 ) = 0.1
That is, ( x - 2.9/0.0424 ) = -1.2816
--> x = -1.2816 * 0.0424 + 2.9 = 2.8457
there 90% chance that sample average is above 2.8547
e.
75% of probability that sample average is below = P ( Z > x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is 0.6745
P( x-u / (s.d) > x - 2.9/0.0424) = 0.25
That is, ( x - 2.9/0.0424) = 0.6745
--> x = 0.6745 * 0.0424+2.9 = 2.9286
there 75% of probability that sample is below 2.9286