Question

1. A recent report states that on a Friday night, the waiting time to be seated at cheesecake factory is normally distributed with a mean of 165 minutes and a standard deviation of 40 minutes. A research agency selects a random sample of 20 people waiting to be seated on a Friday night.

1. What is the standard error of the mean in this sample?
2. What is the likelihood that the sample mean is between 160 and 175 minutes?
3. What is the likelihood that the sample mean is greater than 175 minutes?

Answer 1

Solution :

Given that,

mean = = 165

standard deviation = = 40

n = 20

a) = = 165

= / n = 40 / 20 = 8.94

b) P(160 < < 175)  

= P[(160 - 165) / 8.94 < ( - ) / < (175 - 165) / 8.94)]

= P(-0.56 < Z < 1.12)

= P(Z < 1.12 ) - P(Z < -0.56)

Using z table,  

= 0.8686 - 0.2877

= 0.5809

c) P( > 175) = 1 - P( < 175)

= 1 - P[( - ) / < (175 - 165 ) / 8.94 ]

= 1 - P(z < 1.12 )   

= 1 - 0.8686

= 0.1314