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. 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 $102?

(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 $99 and $100?

(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 = = $99.97  and  Standard deviation = = $18 and  n = 43

(a) Standard error of the mean = = 18/ = 2.74

(b) Let X bar = Sample Mean

The z score probability distribution for sample mean is ;

           Z = ~ N(0,1)

So, Probability that the sample mean will be less than $102 = P(X bar < $102)

  P(X bar < 102) = P( < ) = 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( > ) = 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( < ) = P(Z < 0.01) = 0.50399

    P(X bar <= 99) = P( <= ) = 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 .

2 2/3 hours duhh silly

step-by-step explanation: