Download Chapter 3 Unit Test

Chapter 3 Unit Test
Solutions
Question #1
a. Mean =
5(3)  6( 2)  8(8)  9(7)  10(5)
25
204

 8.16
25
b. Median – 25 numbers / 2= 12.5, so the 13th number is
the median Median = 8
c. Q1 = Median of bottom half = 5 5 5 6 6 8 8 8 8 8 8 8 = 8
d. Standard Deviation
3(5  8.16) 2  2(6  8.16) 2  8(8  8.16) 2  7(9  8.16) 2  5(10  8.16) 2

25
61.36

 1.567
25
Question #2
• Box Plot A – range of 38, IQR of 22, median of 70
• Box Plot B – range of 44, IQR of 21, median of 56
• Box Plot C – range of 60, IQR of 31, median of 65
a. A median of 56 = Box Plot B
b. A range of 60 = Box Plot C
c. Most Consistent = Box Plot B (has smallest IQR)
Question #3
• Mean =  1(3)  2(4)  3(9)  4(10)  5(6)  6(1)
33
 3.45
Question #4
a. The class median would be raised by 3 marks.
The person’s mark that would have been the
middle remains the same, however, it will now
be 3 marks higher.
b. The standard deviation remains the same. The
mean is 3 marks higher, however, every distance
from the mean is the same as before the mark
change … the value under the square root sign
will remain the same, and the number of marks
will be the same.
Question #5
• Z-Score = 1 (1 std. dev) – using table = 84.13%
• Z-Score = 2 (2 std. dev) – using table = 97.72%
97.72% – 84.13% = 13.59%
y
0.25
0.2
0.15
0.1
0.05
x
x 
x  2
x
Question #6
Mean = 5, std dev = 10
X = mean – std dev – std dev
X = 5 – 10 – 10
X = -15
Therefore X would have a value of -15 at -2
standard deviations from the mean.
Question # 7a)
Z-Score of 65 popsicles =
65  80
 2.14
7
Z-Score of 90 popsicles =
90  80
 1.43
7
 1.62%
 93.36%
P(selling 65 to 90) = 93.36% - 1.62% = 90.74%
90.74% * 365 days = 331 days
Therefore, you could expect to sell between 65
and 90 popsicles 331 days of the year.
Question # 7b
• I would think that the sales would follow a
cyclical / yearly pattern (Peaking in the summer).
So the date would be a better prediction (time
series, line graph)
• I would think that sales would be correlated with
temperature (i.e. more sales on hot days). So the
forecast / temperature could also be a good
predictor. (Scatterplot with line of best fit).
• Normal Distribution is not a good predictor in this
case.
Question #8a)
80  72
 0.67  using table = 74.86%
Z-Score =
12
This is the amount below 80%, need to subtract it
from 100 to get the amount above 80%.
100 – 74.86 = 25.14%
Therefore 25.14% of the students are achieving
above the expectations.
Question #8b)
• Q1 = 25% -- using z-score table = -0.67
• Q3 = 75% -- using z-score table = 0.67
Q1 Value
Q1  72
 0.67 
12
 Q  72 
 0.67(12)   1
12
 12 
 8.04  72  Q1  72  72
63.96  Q1
Q3 Value
Q1  72
12
 Q  72 
0.67(12)   1
12
 12 
8.04  72  Q1  72  72
0.67 
80.04  Q1
Therefore, the IQR contains marks from 63.96% to
80.04% which gives a range size of 16.08%
Question #9
Jasmine
Joel
Measures of Central
Tendency:
Mean = 69.7
Mean = 70.4
Std Dev. = 62.69
Std. Dev. = 20.18
Q1 = 20
Q1 = 60
Median = 40
Median = 74.5
-All very close but Joel’s
median is $74.50 compared
to Jasmine’s $40. This
means that 50% of the time
Jasmine earns less than $40
a night. Joel only earned
less than $40 one night.
Q3 = 120
Q3 = 85
IQR = 100
IQR = 25
If tips are a measure of
satisfaction, Joel’s
customers seem a lot more
satisfied than Jasmine’s.
Question #9 - continued
• Joel is more consistent with a standard deviation of $20.12
compared to Jasmine’s $62.70. Similarly, his tips are within
a $25 range (IQR) 50% of the time, where Jasmine’s IQR is
$100.
• Joel’s Q1 is higher than Jasmine’s median, which means
that 75% of the time he earns more than her median of
$40.
See Box Plots on Board
Question #9 - continued
Conclusions: I would have Joel. He is far more
consistent which is important in a restaurant to
ensure customer satisfaction. Although Jasmine
has some great nights ($200, $150) she has many
other nights of adequate satisfaction compared
to Joel.
Limitations: working same nights? Food quality
affects tips? Number of tables? Sections?
(smaller groups = smaller bill = smaller tip) Short
period of time, how accurate a picture?