A survey of 50 retail stores revealed that the average price of a microwave was $375 with a sample standard deviation of $20. Assuming the population is normally distributed, what is the 99% confidence interval to estimate the true cost of the microwave

The 99% C.I is ( 367.703, 382.297 )

Step-by-step explanation:

Given that:

sample mean = 375

Population standard deviation = 20

Sample size n = 50

The level of significance =  1 - C.I = 1 - 0.99 = 0.01

From z tables:

The critical value = = = 2.58

The margin of error E =

The margin of error E =  

The margin of error E = 7.297

The limits of 99% C.I are expressed as:

Lower limit = = 375 - 7.297

Lower limit = 367.703

Upper limit = =  375 + 7.297

Upper limit = 382.297

The 99% C.I is =

= ( 367.703, 382.297 )

594 square inches

step-by-step explanation:

vote !

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.