One college class had a total of 85 students The average sco

One college class had a total of 85 students. The average score for the class on the last exam was 86.6 with a standard deviation of 5.2 A random sample of 33 students was selected.

a. Calculate the standard error of the mean.

The standard error of the mean is (   )?

(Round to two decimal places as needed.)

b. What is the probability that the sample mean will be less than 88?

The probability that the sample mean will be less than 88 is (   )?

(Round to four decimal places as needed.)

c. What is the probability that the sample mean will be more than 87?

The probability that the sample mean will be more than 87 is (   )?

(Round to four decimal places as needed.)

d. What is the probability that the sample mean will be between 85.5 and 87.5?

The probability that the sample mean will be between 85.5 and 87.5 is (   )?

(Round to four decimal places as needed.)

Solution

Mean ( u ) =86.6
Standard Deviation ( sd )=5.2
Number ( n ) = 33
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)
a)
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )=5.2
Sample Size(n)=33
Standard Error = ( 5.2/ Sqrt ( 33) )
= 0.91
                  
b)
P(X < 88) = (88-86.6)/5.2/ Sqrt ( 33 )
= 1.4/0.9052= 1.5466
= P ( Z <1.5466) From Standard NOrmal Table
= 0.939                  
c)
P(X > 87) = (87-86.6)/5.2/ Sqrt ( 33 )
= 0.4/0.905= 0.4419
= P ( Z >0.4419) From Standard Normal Table
= 0.3293                  
d)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 85.5) = (85.5-86.6)/5.2/ Sqrt ( 33 )
= -1.1/0.9052
= -1.2152
= P ( Z <-1.2152) From Standard Normal Table
= 0.11215
P(X < 87.5) = (87.5-86.6)/5.2/ Sqrt ( 33 )
= 0.9/0.9052 = 0.9943
= P ( Z <0.9943) From Standard Normal Table
= 0.83995
P(85.5 < X < 87.5) = 0.83995-0.11215 = 0.7278