According to a research institution, the average hotel price in a certain year w

Question

According to a research institution, the average hotel price in a certain year was $101.34. Assume the population stand of 31 hotels was selected. Complete parts a through d below deviation is $25.00 and that a random sample a. Calculate the standard error of the mearn (Round to two decimal places as needed.) b. Wha P(x $107) = (Round to four decimal places as needed.) d. What is the probability that the sample mean will be between $98 and $101? P($98s- $101)- (Round to four decimal places as needed.) t is the probability that the sample mean will be less than $104?

Explanation / Answer

Mean is 101.34 and standard deviation s is 25, standard error, SE =s/sqrt(N)=25/sqrt(31)=4.49

a) Standard error is 4.49

b) P(xbar<104)=P(z<(104-101.34)/4.49)=P(z<0.59), from normal distribution table we get 0.7224

c) P(xbar>107)=P(z>(107-101.34)/4.49)=P(z>1.26) or 1-P(z<1.26)=1-0.8962 =0.1038

d) P(98<xbar<101)=P((98-101.34)/4.49<z<(101-101.34)/4.49)=P(-0.74<z<-0.08)=P(z<0.74)-P(z<0.08)

from normal distribution table we get 0.7704-0.5319=0.2385

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