Question

Suppose that the Average student is 20.6 years with a standard deviation of 1.7 years. Suppose we obtain a simple random sample of 100 students ages.

1) What is the expected sample average age?

2) What is the standard deviation of the sample average?

3) Find the probability that the sample average is at least 21.3 years.

Answer 1

1)

Expected sample average = = 20.6

2)

Standard deviation of = / sqrt(n)

= 1.7 / sqrt(100)

= 0.17

3)

Using central limit theorem,

P( < x) = P (Z < x - / / sqrt(n) )

So,

P( >= 21.3 ) = P( Z >= 21.3 - 20.6 / 0.17)

= P( Z >= 4.12)

= 0 (from Z table)