1. Which of the following would “most likely” be a better measure of a set of observations
describing the incomes of a population?
A. the mean
B. the median
C. the mode
D. the range
Incomes may be skewed to right or left. So mean isn’t appropriate, median is.
Use the following information to answer questions 2-10:
Over the last two months, I have recorded the lengths of time (in minutes) that I have had to wait
to find a parking space at my local Costco. They are the following:
1.5 .5 2.5 4.0 3.2 2.0 4.3 2.4 1.75 0.0
2. Compute the mean time that I have had to wait to find a parking space over the last two
months.
A. 2.00 minutes
B. 2.46 minutes
C. 2.77 minutes
D. 2.22 minutes
Mean µ = Σx/n = (1.5+.5+2.5+4+3.2+2+4.3+2.4+1.75+0.0)/10 = 2.22
3. Compute the median time that I have had to wait to find a parking space over the last two
months.
A. 2.0 minutes
B. 2.20 minutes
C. 2.22 minutes
D. 2.40 minutes
Arrange in increasing order and take the average of middle two values (5th and 6th value).
0.0, 0.5, 1.5, 1.75, 2, 2.4, 2.5, 3.2, 4, 4.3
Median = (2+2.4)/2 = 2.2
4. The variance of my mean wait time for a parking space at Costco is:
A. 4.15
B. 1.72
C. 1.31
D. 17.24
Variance = (Σ(x-µ)²)/n = ((1.5-2.2)^2+(.5-2.2)^2+(2.5-2.2)^2+(4-2.2)^2+(3.2-2.2)^2+(2-
2.2)^2+(4.3-2.2)^2+(2.4-2.2)^2+(1.75-2.2)^2+(0.0-2.2)^2)/10
= 1.72
5. By the way, how many times have I visited Costco in the last two months?
A. 10
B. 9
C. unknown, because this is a sample of the number of times that I have really visited
D. 8, due to (n-1) degrees of freedom
Count n = 10
6. Using the data above, what is the probability that the next time I visit Costco, I will be able to
find a parking space in 1.5 minutes?
A. 15 percent
B. 1.5 percent
C. 10 percent
D. 90 percent
Probability = Number of times 1.5 occur / Total number of times = 1/10 = 10%
7. The data above can best be described by a:
A. discrete probability distribution
B. continuous probability distribution
C. neither
Time is contiuous.
8. The distribution of my wait time for a parking space at Costco follows a normal distribution.
What is the probability that next time I visit Costco, my wait time for a parking space will be less
than 5 minutes?
A. 100%
B. 98%
C. 48%
D. 2.12%
P(x<5)=P(z < (x-µ)/σ) = P(z < (5-2.22)/sqrt(1.72)) = P(z<2.12) = 98%
9. The distribution of my wait time for a parking space at Costco follows a normal distribution.
What is the probability that next time I visit Costco, my wait time for a parking space will be
between 2.22 and 4 minutes?
A. .4131
B. .0900
C. .9100
D. .0136
Z for 2.22 = (2.22-2.22)/sqrt(1.72) = 0
Z for 4 = (4-2.22)/sqrt(1.72) = 1.357
P(0<z<1.357) = P(z<1.357)-P(z<0) = 0.913 – 0.5 = 0.413
10. The distribution of my wait time for a parking space at Costco follows a normal distribution.
What is the probability that next time I visit Costco, my wait time for a parking space will be
between .5 and 4 minutes?
A. 41 percent
B. 40 percent
C. 99 percent
D. 81 percent
Z for 0.5 = (0.5-2.22)/sqrt(1.72) = -1.312
Z for 4 = (4-2.22)/sqrt(1.72) = 1.357
P(-1.312<z<1.357) = P(z<1.357) – P(z<-1.312) = 0.913 – 0.095 = 0.81 = 81%
Open the following link: www.ual.com and enter fictitious information for a round trip flight of
your choice. When you are given several itinerary choices, review the operation trends under
“flight details” for your departure or return and answer questions #11-12.
11. These operational trends were most likely calculated using a:
A. Poisson distribution
B. Standard normal distribution
C. Binomial probability distribution
D. Relative frequency distribution
It would have been obtained from actual data without assuming any distribution.
12. Which of the following would you most likely use to calculate the probability that exactly 2
of the many departures offered on your chosen date would arrive late, given the probability that
any of the departing flights arriving late is .08?
A. Poisson distribution
B. Standard normal distribution
C. Binomial probability distribution
D. Relative frequency distribution
Binomial probability as there are only two possibilities, late or not late, also probability of
each flight being late is same.
13. The wildlife department has been feeding a special food to rainbow trout fingerlings in a
pond. A sample of the weights of 40 trout revealed that the mean weight is 402.7 grams and the
standard deviation 8.8 grams. What is the probability that the mean weight for a sample of 40
trout exceeds 405.5 grams?
A. 0.3783
B. 0.0228
C. 1.0
D. 0.5
Distribution of ‘Means’ has standard error = 8.8/sqrt(40) = 1.3914
Mean value of distribution remains same = 402.7
So, P(x>405.5) = P(z > (405.5-402.7)/1.3914) = P(z>2.012) = 0.022
14. For a distribution of sample means constructed by sampling 5 items from a population of 15,
A. the sample size is 15
B. there will be 3003 possible sample means
C. the mean of the sample means will be 3
D. the standard error will be 1
Possible number of samples chosen = 15C5 = 3003
15. An accounting firm is planning for the next tax preparation season. From last year’s returns,