According to a research institution the average hotel price

According to a research institution, the average hotel price in a certain year was $99.21. Assume the population standard deviation is $18.00 and that a random sample of 37 hotels was selected. Complete parts (a) through (d) below.

a. Calculate the standard error of the mean.

(Round to two decimal places as needed.)

b. What is the probability that the sample mean will be less than $100?

(Round to four decimal places as needed.)

c. What is the probability that the sample mean will be more than $104?

(Round to four decimal places as needed.)

d. What is the probability that the sample mean will be between $98 and $99?

(Round to four decimal places as needed.)

Solution

a)
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )=18
Sample Size(n)=37
Standard Error = ( 18/ Sqrt ( 37) )
= 2.96
b)
P(X < 100) = (100-99.21)/18/ Sqrt ( 37 )
= 0.79/2.9592= 0.267
= P ( Z <0.267) From Standard NOrmal Table
= 0.6053                  
c)
P(X > 104) = (104-99.21)/18/ Sqrt ( 37 )
= 4.79/2.959= 1.6187
= P ( Z >1.6187) From Standard Normal Table
= 0.0528                  
d)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 98) = (98-99.21)/18/ Sqrt ( 37 )
= -1.21/2.9592
= -0.4089
= P ( Z <-0.4089) From Standard Normal Table
= 0.34131
P(X < 99) = (99-99.21)/18/ Sqrt ( 37 )
= -0.21/2.9592 = -0.071
= P ( Z <-0.071) From Standard Normal Table
= 0.47171
P(98 < X < 99) = 0.47171-0.34131 = 0.1304