Test 1 Answers
1. A city police officer, using radar, checked the speed (in mph) of twelve cars as they were traveling down
a city street. The table below shows some of the statistics. Address parts (a) and (b).
statistic
mean
median
1st Quartile
3rd Quartile
standard deviation
value
30.0
29.0
23.0
34.5
9.6
(a) If a variable’s value is more than 1.5 ⋅ IQR units from either end of the box in a boxplot, then it is
labeled an outlier. Use this definition of an outlier to determine if the label outlier can be applied
to the car the police officer measured traveling 53 mph. Justify your answer.
We must determine whether 53 is more than 1.5 ⋅ ( IQR ) units below Q 1 or more than 1.5 ⋅ ( IQR ) units
above Q 3. That is, we must determine whether 53 is larger than the Upper Fence.
Recall: Upper Fence = Q3 + 1.5 IQR and IQR = Q3 − Q1 .
Thus, in this particular problem, IQR = Q3 − Q1 = 34.5 − 23.0 = 11.5 (where the values of Q 1 and Q 3
were given). And so, 1.5 ⋅ ( IQR ) = 1.5 ⋅ (11.5 ) = 17.25 .
Since Lower Fence = Q1 − 1.5 ⋅ ( IQR ) = 23.0 − 17.25 = 5.75
and Upper Fence = Q3 + 1.5 ⋅ ( IQR ) = 34.5 + 17.25 = 51.75 ,
then any data value less than 17.25 and any data value greater than 51.75 would be classified as an outlier.
Since the value 53 satisfies this criteria (it is greater than 51.75), then 53 is labeled as an outlier.
Answer: Yes, the police officer can treat the 53 mph car speed as an outlier since it is more than
1.5 ⋅ IQR s outside this data set’s boxplot’s box (specifically, it is greater than 1.5 ⋅ ( IQR )
units above Q 3).
(b) One car's speed is 1¼ standard deviations below the twelve cars' mean speed; determine this particular car's
speed (in mph). Justify your answer.
Since 1 standard deviation equals 9.6 mph, then 1¼ standard deviations equals ( 1¼ )( 9.6 ) = 12 mph .
Thus, the speed that is 1¼ standard deviations below this sample’s mean speed of 30 mph is 30 – 12
= 18 mph.
Alternatively, since the car's speed is 1¼ standard deviations below the twelve cars' mean speed, then (by
definition) it's z-score is −1.25 .
x − mean
takes on numerous forms depending on whether you
std dev
are working with a population, a sample, or a sampling distribution. Since we are working with a sample
x−x
here, then the formula takes on the form z =
, where the sample's mean is designated by x and the
s
sample's standard deviation is denoted by s.
Recall that the generic formula z =
Thus, with z = −1.25 , x = 30.0 ,and s = 9.6 , then z =
−1.25 =
x − 30.0
⇒
9.6
( 9.6 )( −1.25) = x − 30.0
x−x
becomes
s
⇒ − 12 = x − 30.0 ⇒ − 12 + 30.0 = x ⇒ 18 = x .
Answer: The car’s speed is 18 mph.
2. The amount of time customers must wait for service at the drive-thru window of the Marysville Bank of America
is Normally Distributed with mean 9.2 minutes and standard deviation 2.6 minutes. Address parts (a) & (b).
Please specify the calculator function you use; include all of the values you enter into your calculator for each.
Report each answer with two digits after the decimal point. Sentence answers not required.
(a) Determine that the percentage of all this bank's drive-thru customers that wait less than 8 minutes for
service at the drive-thru window.
With x representing the service time (in minutes) of ALL the customers who use the drive-thru window
of the Marysville Bank of America, we have been given that the distribution of x is Normal with mean
µ = 9.2 minutes and standard deviation σ = 2.6 minutes.
We are asked to find P ( x < 8 ) . Since the smallest possible value of x is 0 and since x is a continuous
random variable (and so including the 8 will not change the amount of area determined), we want the
value of P ( 0 ≤ x ≤ 8 ) .
P ( 0 ≤ x ≤ 8 ) = normalcdf ( 0, 8, 9.2, 2.6 )
≈ 0.3220049068
≈ 32.20% [rounded as instructed]
0
8
Answer: Approximately 32.20% of all this bank's drive-thru customers wait less than 8 minutes for
service at the drive-thru window.
(b) Determine the first quartile, Q1.
Q1 is the value of the variable that separates the lower 25% of the variable’s values from the upper 75%.
Thus, Q1 = invNorm ( 0.25, 9.2, 2.6 ) ≈ 7.446326651 ≈ 7.45 [rounded as instructed].
Answer: Q1 ≈ 7.45 minutes
3. The scores of the top ten finishers in a recent Ladies Professional Golf Association (LPGA) Valley of the Stars
Tournament are listed below.
71, 62, 62, 77, 76, 72, 75, 68, 71, 71
Find the median score.
-------------------------------------------------------------------------------------------------------------------------------------------Enter the ten numbers into one of the lists in the calculator, say L1, then use 1-VarStats L1. The median
score is 71.
Answer: 71 golf score
4. The heights (in inches) of ten adult males are listed below. These data values were obtained from a sample of
American men.
70, 73, 70, 74, 65, 73, 69, 68, 70, 71
Find the standard deviation.
-------------------------------------------------------------------------------------------------------------------------------------------Enter the ten numbers into one of the lists in the calculator, say L1, then use 1-VarStats L1. The sample, versus
population standard deviation (σ), is s ≈ 2.7 inches.
Answer: 2.7 inches*
--------------------------------------------------------------------------------------------------------------------------------------------------* The Rounding Rule of Statistics states that calculations computed using sample data should be reported with one
more digit than the raw data has (www.tc3.edu/instruct/sbrown/stat/rounding.htm or
www.saddleback.edu/faculty/pquigley/math10/rounding.pdf)
5. Data pertaining to the number of cars per household was collected on a sample of 1000 Seattle households last year.
The responses are listed in the table.
Cars
0
1
2
3
4
Households
125
464
216
123
72
Determine the mean number of cars per household.
Enter the values of the variable into one of the lists in the calculator, say L1, and enter the corresponding number
of households (i.e. the frequency of each value of the variable) into a second list in the calculator, say L2. Then
use 1-VarStates L1, L2. The mean number of cars is 1.6 cars (reported with the number of digits dictated by the
Rounding Rule of Statistics).
Answer: 1.6 cars
6. Data is collected on daily temperatures ( in ° F ) recorded in Nome, Alaska during the winter months. These
temperatures were recorded to the nearest whole Integer and the following histogram of this sample’s data was
constructed.
Compute the variance of this data set. Report your answer with at least three digits after the decimal point.
Extracting the data from the frequency histogram and putting it into a frequency table, we have:
Temperatures ( in °F ) in
Nome, Alaska
−5
−4
−3
−2
−1
0
1
2
3
4
5
Frequency
(Number of Days)
1
2
3
4
5
6
5
4
3
2
1
Enter the values of the variable into one of the lists in the calculator, say L1, and enter the corresponding
frequency of each value of the variable into a second list in the calculator, say L2. Then use 1-VarStates L1,
L2. Since this data was collected on a sample, then the standard deviation is s or s x . Recalling this unrounded value from within the calculator’s VARS menu by choosing the Statistics option (the 5th selection),
2
and squaring it, we find that the variance is s 2 or sx 2 ≈ ( 2.449489743 ) = 6 . Thus, reported with three
digits after the decimal point, the variance is 6.000 ° F 2 .
Answer: 6.000 ° F 2
7. The table depicts the number of people, amongst 279 French skiers studied, who contracted a common
colds during ski season. Unbeknownst to either group, some of the skiers took a sugar pill every morning (this
“fake” medication is referred to as a placebo); the remaining skiers took a cold treatment pill every morning.
Placebo
(Sugar Pill)
Treatment
(Cold Treatement Pill)
Totals
Cold
No Cold
Totals
31
109
140
17
122
139
48
231
279
Compute each of the following. No work or explanation is necessary; sentence answers not required.
(a) The proportion of the skiers who took a sugar pill every day and didn’t get a cold?
Answer:
109
279
(b) The proportion of those skiers who didn’t get a cold that took a sugar pill every day.
Answer:
109
231
(c) The proportion of the skiers that either received the treatment or who didn’t get a cold.
Answer:
248
279