Respuesta :
Answer:
(a) 2.74
(b) 0.7704
(c) 0.0708
(d) 0.14082
Step-by-step explanation:
We are given that according to a research institution, the average hotel price in a certain year was $99.97.Assume the population standard deviation is $18.00 and that a random sample of 43 hotels was selected.
Here, Mean = [tex]\mu[/tex] = $99.97 and Standard deviation = [tex]\sigma[/tex] = $18 and n = 43
(a) Standard error of the mean = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = 18/[tex]\sqrt{43}[/tex] = 2.74
(b) Let X bar = Sample Mean
The z score probability distribution for sample mean is ;
Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
So, Probability that the sample mean will be less than $102 = P(X bar < $102)
P(X bar < 102) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{102-99.97}{\frac{18}{\sqrt{43} } }[/tex] ) = P(Z < 0.74) = 0.7704
(c) Probability that the sample mean will be more than $104 =P(X bar > $104)
P(X bar > 104) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{104-99.97}{\frac{18}{\sqrt{43} } }[/tex] ) = P(Z > 1.47) = 1 - P(Z <= 1.47)
= 1 - 0.92922 = 0.0708
(d) Probability that the sample mean will be between $99 and $100 is given by = P($99 < X bar < $100) = P(X bar < $100) - P(X bar <= $99)
P(X bar < 100) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{100-99.97}{\frac{18}{\sqrt{43} } }[/tex] ) = P(Z < 0.01) = 0.50399
P(X bar <= 99) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{99-99.97}{\frac{18}{\sqrt{43} } }[/tex] ) = P(Z <= -0.35) = 1 - P(Z <= 0.35)
= 1 - 0.63683 = 0.36317
Therefore, P($99 < X bar < $100) = 0.50399 - 0.36317 = 0.14082 .
