Question

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. The distribution of interpupillary distance is unknown.

a.Suppose that a sample of size 100 is taken from this population. What is the expected value of the sample mean? The standard deviation (called standard error) of the mean? What distribution does the sample mean follow? Justify your answers.

What is the probability that a randomly selected sample of size 100 will have an average interpupillary distance between 64 and 66 mm?

Answer 1

Solution :

Given that,

mean = = 65

standard deviation = = 5

n = 100

= = 65

= / n = 5 / 100 = 0.5

The distribution of sample mean is approximately normal because n is greater than 30, so use Central Limit Theorem.

P(64 < < 66 )  

= P[(64 - 65) / 0.5 < ( - ) / < (66 - 65) / 0.5 )]

= P(-2.00 < Z < 2.00 )

= P(Z < 2.00) - P(Z < -2.00)

Using z table,  

= 0.9772 - 0.0228

= 0.9544