Question

You are contacted by a phone-in technical support business that is interested in some information about the amount of time their customers spend on hold. You find out that on average, each caller spends 11 minutes on hold with a standard deviation of 1.16 minutes. If you were to take a random sample of 62 callers, you would expect 79% of the time the average hold time would be greater than how many minutes?

Question 10 options:

	1) 10.88
	2) 11.94
	3) 10.06
	4) There is not enough information to determine this.
	5) 11.12

Fill in the blank. In a drive thru performance study, the average service time for McDonald's is 186.69 seconds with a standard deviation of 6.26 seconds. A random sample of 82 times is taken. There is a 43% chance that the average drive-thru service time is greater than ________ seconds.

Question 11 options:

	1) 186.81
	2) 187.79
	3) 185.59
	4) 186.57
	5) There is not enough information to determine this.

Answer 1

Solution :

10 )Given that,

mean = = 11

standard deviation = = 1.16

n =62

= 11

₌ / n = 1.16 62 = 0.1473

P( Z > z) =79%

P(Z > z) = 0.79

1 - P( Z < z) = 0.79

P(Z < z) = 1 - 0.79

P(Z < z) = 0.21

z = - 0.81

Using z-score formula,

= z * +

= - 0.81 * 0.1473 + 11

= 10.88

Option 10.88 is correct.

11 ) Given that,

mean = = 186.69

standard deviation = = 6.26

n = 82

= 1200

₌ / n = 6.2682 = 0.6913

P( Z > z) = 43%

P(Z > z) = 0.43

1 - P( Z < z) = 0.43

P(Z < z) = 1 - 0.43

P(Z < z) = 0.57

z = 0.18

Using z-score formula,

= z * +

= 0.18 * 0.6913 + 186.69

= 186.81

Option 186.81 is correct.