Question

One college class had a total of 80 students. The average score for the class on the last exam was 84.6 with a standard deviation of 5.3. A random sample of 34 students was selected.

a. Calculate the standard error of the mean.

b. What is the probability that the sample mean will be less than 86​?

c. What is the probability that the sample mean will be more than 85​?

d. What is the probability that the sample mean will be between 83.5 and 85.5​?

Answer 1

Solution :

Given that ,

mean = = 84.6

standard deviation = = 5.3

n = 34

= 84.6

(a) standard error = = / n = 5.3 / 34 = 0.9089

(b)P( >86 ) = 1 - P( < 86)

= 1 - P[( - ) / < (86 - 84.6) /0.9089 ]

= 1 - P(z <1.54 )

Using z table,    

= 1 - 0.9382

= 0.0618

(c)P( >85 ) = 1 - P( < 85)

= 1 - P[( - ) / < (85 - 84.6) /0.9089 ]

= 1 - P(z < 0.44)

Using z table,    

= 1 - 0.6700

= 0.3300

(d)P(83.5< <85.5 )  

= P[(83.5 - 84.6) /0.9089 < ( - ) / < (85.5 - 84.6) /0.9089 )]

= P( -1.21< Z < 0.99)

= P(Z < 0.99) - P(Z < -1.21)

Using z table,  

=   0.8389 - 0.1131

=0.7258