Question

4. Another application of the sampling distribution of the sample mean Suppose that, out of a total of 431 full-service restaurants in Alaska, the number of seats per restaurant is normally distributed with mean µ = 77.9 and standard deviation σ = 20.2. (Source: U.S. Census Bureau; data based on the 2002 economic census). The Alaska tourism board selects a simple random sample of 50 full-service restaurants located within the full-service restaurants located within the state and determines the mean number of seats per restaurant for the sample. The standard deviation of the sampling distribution of the sample mean is ________.

a. 20.2

b. 0.9413

c. 2.86

d. 2.69

There is a .25 probability that the sample mean is less than _______.

a. 75.92

b. 75.99

c. 77.9

d. 76.09

Answer 1

Solution :

a)

The sampling distribution of standard deviation is ,  

= / n = 20.2 / 50 = 2.86

b)

P(Z < -0.67) = 0.25

z = -0.67

= z * + = -0.67 * 2.86 + 77.9 = 75.99

There is a .25 probability that the sample mean is less than  75.99