Question

1.) Years of scores have indicated that the mean on a stats exam is 30 and the standard deviation is 7. The number of people in the sample class 47.

a- What is the population mean of the sampling distribution?

b-What is the population variance and standard error of this sampling distribution?

c-What minimum sample mean would this class need to obtain to be in the top 5% of the sample mean ?

d) What is the probability of obtaining a sample mean of 28.2 or bellow for a class of 49 students?

e) What is the probability that an individual student in class would obtain a score of 28.2 or below?

Answer 1

Solution,

Given that,

mean = = 30

standard deviation = = 7

n = 47

a) = = 30

b) variance = ² = 7² = 49

= / n = 7 / 47 = 1.021

c) Using standard normal table,

P(Z > z) = 5%

= 1 - P(Z < z) = 0.05  

= P(Z < z ) = 1 - 0.05

= P(Z < z ) = 0.95

= P(Z < 1.645 ) = 0.95  

z = 1.645

Using z-score formula  

= z * +

= 1.645 *1.021 + 30

= 31.68

d) P( 28.2) = P(( - ) / (28.2 - 30) / 1.021)

= P(z -1.76)

Using z table

= 0.0392

e) P(x 28.2)

= P[(x - ) / (28.2 - 30) / 7]

= P(z -0.26)

Using z table,

= 0.3974