Question

Waiters at a restaurant chain earn an average of $249 per shift (regular pay + tips) with a standard deviation of $39. For random samples of 36 shifts at the restaurant, within what range of dollar values will their sample mean earnings fall, with 95% probability?

Standard Normal Distribution Table

Range:

to

Round to the nearest cent

Answer 1

solution:

Given that,

mean = = 249

standard deviation = = 39

n = 36

= 249

= / n = 39/36=6.5

middle % of score is

P(-z < Z < z) = 0.95

P(Z < z) - P(Z < -z) = 0.95

2 P(Z < z) - 1 = 0.95

2 P(Z < z) = 1 + 0. 95= 1.95

P(Z < z) = 1.95/ 2 = 0.975

P(Z <1.96 ) = 0.975

z  ±1.96   

Using z-score formula  

= z * +   

= ±1.96*6.5+249

=236.26 to  261.74