Suppose z is the standard normal random variable. Use Excel to ...

[frankwu]

Suppose z is the standard normal random variable. Use Excel to calculate the following:
 a. P(z  2.5)
  b. P(0  z  2.5)
  c. P(2  z  2)
  d. P(z  0.38)
  e. P(z  1.62)
  f. z value with 0.05 in the lower tail
  g. z value with 0.05 in the upper tail

Question 2

Fifty students are enrolled in an Economics class. After the first examination, a random sample of five papers was selected. The grades were 60, 75, 80, 70, and 90.
 a. Calculate the estimate of the standard error of the mean.
  b. What assumption must be made before we can determine an interval for the mean grade of all the students in the class? Explain why.
  c. Assume the assumption of part (b) is met. Provide a 90 confidence interval for the mean grade of all the students in the class.
  d. If there were 200 students in the class, what would be the 90 confidence interval for the mean grade of all the students in the class?

[ciecieme]

Answer to Question 1

a. P(z  2.5) =NORM.S.DIST(2.5,TRUE)
b. P(0  z  2.5) =NORM.S.DIST(2.5,TRUE)-NORM.S.DIST(0,TRUE)
c. P(2  z  2) =NORM.S.DIST(2,TRUE)-NORM.S.DIST(-2,TRUE)
d. P(z  0.38) =NORM.S.DIST(-0.38,TRUE)
e. P(z  1.62) =1-NORM.S.DIST(1.62,TRUE)
f. z value with 0.05 in the lower tail =NORM.S.INV(0.05)
g. z value with 0.05 in the upper tail =NORM.S.INV(0.95)

Answer to Question 2

a. 4.79
b. Since the sample is small (n < 30) and  is estimated from s, we must assume the distribution of all the grades is normal.
c. 64.783 to 85.217
d. 64.34 to 85.66

[frankwu]

Wow, this really help

[parker125]

Thanks for the timely response, appreciate it