Question

One college class had a total of

8080

students. The average score for the class on the last exam was

83.983.9

with a standard deviation of

5.85.8.

A random sample of

3232

students was selected.

a. Calculate the standard error of the mean.

b. What is the probability that the sample mean will be less than

8585​?

c. What is the probability that the sample mean will be more than

8484​?

d. What is the probability that the sample mean will be between

82.582.5

and

84.584.5​?

a. The standard error of the mean is

nothing.

​(Round to two decimal places as​ needed.)

b. The probability that the sample mean will be less than

8585

is

nothing.

​(Round to four decimal places as​ needed.)

c. The probability that the sample mean will be more than

8484

is

nothing.

​(Round to four decimal places as​ needed.)

d. The probability that the sample mean will be between

82.582.5

and

84.584.5

is

nothing.

​(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 83.9

standard deviation = = 5.8

n = 32

= = 83.9

a) = / n = 5.8 / 32 = 1.03

b) P( < 85) = P(( - ) / < (85 - 83.9) / 1.03)

= P(z < 1.07)

Using z table

= 0.8577  

c) P( > 84) = 1 - P( < 84)

= 1 - P[( - ) / < (84 - 83.9) / 1.03 ]

= 1 - P(z < 0.10)

Using z table,    

= 1 - 0.5398

= 0.4602

d) P(82.5 < < 84.5)  

= P[(82.5 - 83.9) /1.03 < ( - ) / < (84.5 - 83.9) / 1.03)]

= P(-1.36 < Z < 0.58)

= P(Z < 0.58) - P(Z < -1.36)

Using z table,  

= 0.7190 - 0.0869

= 0.6321