Question

Waiters at a restaurant chain earn an average of $256 per shift (regular pay + tips) with a standard deviation of $35. For random samples of 36 shifts at the restaurant, within what range of dollar values will their sample mean earnings fall, with 98% probability?

Range: ? to ?

Answer 1

Solution,

Given that,

mean = = 256

standard deviation = = 35

n = 36

= = 256

= / n = 35 / 36 = 5.83

Using standard normal table,

P( -z < Z < z) = 98%

= P(Z < z) - P(Z <-z ) = 0.98

= 2P(Z < z) - 1 = 0.98

= 2P(Z < z) = 1 + 0.98

= P(Z < z) = 1.98/ 2

= P(Z < z) = 0.99

= P(Z < 2.33 ) = 0.99

= z  ± 2.33

Using z-score formula  

= z * +

= -2.33 * 5.83 + 256

= 242.4

Using z-score formula  

= z * +

= 2.33 * 5.83 + 256

= 269.6

Range = 242.4 to 269.6