Question

The weight of bags of organic fertilizer is normally distributed with a mean of 50 pounds and a standard deviation of 1 pound. If we take a random sample of 25 bags of organic fertilizer, there is an 80% chance that their mean weight will be greater than what value?

Suppose that IQs of adult Canadians follow a normal distribution with standard deviation 15. A random sample of 30 adult Canadians has a mean IQ of 112. We would like to construct a 92% confidence interval for the true mean IQ of all adult Canadians. What is the critical value z* to be used in the interval? (Input a positive number since we always use the positive z* value when calculating confidence intervals.) Report your answer to 2 decimal places.

Answer 1

a) P( > x) = 0.8

or, P(( - )/() > (x - )/()) = 0.8

or, P(Z > (x - 50)/(1/)) = 0.8

or, P(Z <  (x - 50)/(1/)) = 0.2

or,   (x - 50)/(1/) = -0.84

or, x = -0.84 * (1/) + 50

or, x = 49.832

b) At 92% confidence interval the critical value is z^* = 1.75