Question

The amount of time a bank teller spends with each customer has a population​ mean, μ​, of 2.90 minutes and a standard​ deviation, σ​, of 0.40 minute. Complete parts​ (a) through​ (c).

a. If you select a random sample of 16 customers, what is the probability that the mean time spent per customer is at least 2.7 minutes?

nothing

​(Round to four decimal places as​ needed.)

b. If you select a random sample of 16 customers, there is an 84​% chance that the sample mean is less than how many​ minutes?

​(Round to four decimal places as​ needed.)

c. If you select a random sample of 100 customers, there is an 84​% chance that the sample mean is less than how many​ minutes?

(Round to four decimal places as​ needed.)

Answer 1

Solution :

Given that,

mean = = 2.90

standard deviation = = 0.40

n = 16

= = 2.90

= / n = 0.40 / 16 = 0.1

a) P( 2.7) = 1 - P( 2.7)

= 1 - P[( - ) / (2.7 - 2.90) / 0.1 ]

= 1 - P(z -2.0)

Using z table,    

= 1 - 0.0228

= 0.9772

b) Using standard normal table,

P(Z < z) = 84%

= P(Z < z ) = 0.84

= P(Z < 0.99 ) = 0.84  

z = 0.99

Using z-score formula  

= z * +   

= 0.99 * 0.1 + 2.90

= 2.999

c) n = 100

= = 2.90

= / n = 0.40/ 100 = 0.04

Using standard normal table,

P(Z < z) = 84%

= P(Z < z ) = 0.84

= P(Z < 0.99) = 0.84  

z = 0.99

Using z-score formula  

= z * +   

= 0.99 * 0.04 +2.90

= 2.9396