(1) Hart Hotel is a large hotel catering to business and vacation travelers. The
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Question
(1) Hart Hotel is a large hotel catering to business and vacation travelers. The hotel has fifty suites, each one named for a State: the Alabama Suite, the Alaska Suite, etc. The daily charge for a suite depends on its size [in square feet] and which floor of the hotel it occupies. Each suite has unlimited free internet access, but guests do pay for phone calls, with the charge based on the length of the call. Frequent guests are eligible for one of four discount plans, which are, from lowest to highest: Traveler Discount, Business Discount, International Discount, SuperStar Discount. The number of occupied rooms varies with the season, being the highest during summer and major holidays. Aside from staffing, the hotel’s largest expense item is daily water usage—water consumed by guests and in the hotel laundry.
From the paragraph above:
Which one of the following is a discrete numerical random variable?
Daily water usage.
Number of occupied rooms.
Suite name.
Length of a phone call.
Which one of the following is a continuous numerical random variable?
Daily water usage.
Suite name.
Number of occupied rooms.
Hotel floor.
(c)Which one of the following is a categorical random variable?
Number of occupied rooms.
Length of a phone call.
Suite name.
Daily water usage.
(d)Which one of the following is an ordinal scaled variable?
Length of a phone call.
Number of rooms occupied.
Daily water usage.
Discount plan.
[16 points]
(2) A random sample of four guests’ Hart Hotel bills is given below:
385
1575
2985
3855
(a) Find the arithmetric mean.
2280.
2933.
2200.
None of the above.
(b) Find the median.
2280.
2200.
3470.
None of the above.
Find the range.
2280.
3470.
1410.
None of the above.
(d) Find the sample standard deviation. Round off to the nearest dollar.
1327.
2,346,700.
1532.
None of the above.
[6 points]
(3) Hart Hotel receives substantial revenue from renting its ballroom for a variety of different occasions. The Pareto diagram below shows all ballroom rentals for the year 2012. Use the Pareto diagram to estimate:
What percent of the occasions were weddings?
90%
25%
35%
45%
Which occasions accounted for approximately 85% of all ballroom rentals in 2012?
Bar mitzvah, corporate meeting, & anniversary.
Wedding & anniversary.
Wedding, bar mitzvah, & corporate meeting.
Corporate meeting & anniversary.
[8 points]
(4) According to Hart Hotel records, 80% of all guests checking in have luggage.
In a random sample of 15 guests, what is the probability that 10 or more guests will have luggage? Use the binomial table below to answer.
93.9%
10.3%
83.6%
16.4%
Data
Sample size
15
Probability of success
0.8
Statistics
Mean
12
Variance
2.4
Standard deviation
1.549193
X
P(X)
P(<=X)
P(<X)
P(>X)
P(>=X)
8
0.013819
0.018059
0.00424
0.981941
0.99576
9
0.042993
0.061051
0.018059
0.938949
0.981941
10
0.103182
0.164234
0.061051
0.835766
0.938949
11
0.187604
0.351838
0.164234
0.648162
0.835766
12
0.250139
0.601977
0.351838
0.398023
0.648162
13
0.230897
0.832874
0.601977
0.167126
0.398023
14
0.131941
0.964816
0.832874
0.035184
0.167126
15
0.035184
1
0.964816
0
0.035184
(b) If each guest is an independent event, what is the probability that the next ten guests to check in all have luggage? Round off to the nearest whole percent.
80%
16%
8%
11%
[8 points]
(5) The potholes on a major Chicago highway occur at an average rate of 3.4 per mile. Suppose one mile of highway is randomly selected. Use the Poisson table below to answer the following questions:
(a)What is the probability of more than 1 pothole occurring?
14.68%
19.29%
85.32%
96.66%
(b)What is the probability of fewer than 3 potholes occurring?
21.86%
33.97%
44.16%
85.32%
POTHOLES
Data
Average/Expected number of successes:
3.4
Poisson Probabilities Table
X
P(X)
P(<=X)
P(<X)
P(>X)
P(>=X)
0
0.033373
0.033373
0.000000
0.966627
1.000000
1
0.113469
0.146842
0.033373
0.853158
0.966627
2
0.192898
0.339740
0.146842
0.660260
0.853158
3
0.218617
0.558357
0.339740
0.441643
0.660260
4
0.185825
0.744182
0.558357
0.255818
0.441643
5
0.126361
0.870542
0.744182
0.129458
0.255818
6
0.071604
0.942147
0.870542
0.057853
0.129458
[8 points]
(6) Salmon weights are normally distributed with a mean of 12.3 pounds and a standard deviation of 2 pounds.
What is the probability that a randomly selected salmon weighs between 10 and 14 pounds?
67.72%
32.28%
80.23%
85%
An “Emperor” salmon is one that weighs more than 99% of all salmon. How much must a salmon weigh in order to earn “Emperor” status?
7.64 pounds.
14.86 pounds.
16.96 pounds
Cannot be determined from the information given.
[8 points]
(7) Salmon weights are normally distributed with a mean of 12.3 pounds and a standard deviation of 2 pounds.
(a) What is the mean and standard deviation (the standard error) of the sampling distribution for samples of size 4?
Mean = 12.3 pounds, standard error = 2 pounds.
Mean = 12.3 pounds, standard error = 1 pound.
Mean = 6.15 pounds, standard error = 2 pounds.
Cannot be determined from the information given.
(b) Suppose we increase the sample size from part (a). How will the mean and standard error of the sampling distribution change?
(A) Both will decrease.
(B) The mean will increase, the standard error will decrease.
(C) The mean will stay the same, the standard error will decrease.
(D) Both will stay the same.
385
1575
2985
3855
Explanation / Answer
1.a)No of occupied rooms is a dicrete random variable
b) continuous random variable is daily water usage
c) categorical random variable is suite name
d) A ordinal variable, is one where the order matters but not the difference between values. ... A interval variable is a measurement where the difference between two values is meaningful.
length of a phone call is best one
2) a) mean = (385+1575+2985+3855)/4 =8800/4 = 2200
b) median = 1575+2985 /2 = 4560/2 = 2280
c) range = 3855-385 = 3470
d) standard deviation = 1531.89425 = 1532
3) no pareto diagram
4) a) P(X>=10) = 0.9389 = 93.9%
b)as all ten guests are independent then we have 0.810 = 0.10737 = 10.737% = 11%
5)
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