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```BSTAT 3321 Assignment 2
20 points each question
1. Problem 3.62 a, b, and d
Excel output:
Time
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
First Quartile
Third Quartile
interquartile range
c.v
(a)
mean = 43.89
(b)
range = 76
43.88889
4.865816
45
17
25.28352
639.2564
-0.90407
0.517268
76
16
92
1185
27
18
63
45
57.61%
median = 45
interquartile range = 45
standard deviation = 25.28
(d)
1st quartile = 18 3rd quartile = 63
variance = 639.2564
coefficient of variation = 57.61%
The mean approval process takes 43.89 days with 50% of the policies being approved in
less than 45 days. 50% of the applications are approved between 18 and 63 days.
About 67% of the applications are approved between 18.6 to 69.2 days.
2. Problem 3.64 a, b, and d
Excel output:
Width
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
First Quartile
Third Quartile
Interquartile Range
CV
8.420898
0.006588
8.42
8.42
0.046115
0.002127
0.035814
-0.48568
0.186
8.312
8.498
412.624
49
8.404
8.459
0.055
0.55%
(a)
mean = 8.421, median = 8.42, range = 0.186 and standard deviation = 0.0461. On average, the
width is 8.421 inches. The width of the middle ranked observation is 8.42. The difference between the
largest and smallest width is 0.186 and majority of the widths fall between 0.0461 inches around the mean
of 8.421 inches.
(b)
Minimum = 8.312, 1st quartile = 8.404, median = 8.42, 3rd quartile = 8.459 and maximum = 8.498
(d)
All the troughs fall within the limit of 8.31 and 8.61 inches.
3. Problem 3.67 a, b, c, and e
Excel output:
Teabags
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
First Quartile
Third Quartile
Interquartile Range
CV
5.5014
0.014967
5.515
5.53
0.10583
0.0112
0.127022
-0.15249
0.52
5.25
5.77
275.07
50
5.44
5.57
0.13
1.9237%
(a)
mean = 5.5014, median = 5.515, first quartile = 5.44, third quartile = 5.57
(b)
range = 0.52, interquartile range = 0.13, variance = 0.0112,
standard deviation = 0.10583, coefficient of variation = 1.924%
(c)
The mean weight of the tea bags in the sample is 5.5014 grams while the middle ranked weight is
5.515. The company should be concerned about the central tendency because that is where the majority
of the weight will cluster around. The average of the squared differences between the weights in the
sample and the sample mean is 0.0112 whereas the square-root of it is 0.106 gram. The difference
between the lightest and the heaviest tea bags in the sample is 0.52. 50% of the tea bags in the sample
weigh between 5.44 and 5.57 grams. According to the empirical rule, about 68% of the tea bags produced
will have weight that falls within 0.106 grams around 5.5014 grams. The company producing the tea
bags should be concerned about the variation because tea bags will not weigh exactly the same due to
various factors in the production process, e.g. temperature and humidity inside the factory, differences in
the density of the tea, etc. Having some idea about the amount of variation will enable the company to
(e)
On average, the weight of the teabags is quite close to the target of 5.5 grams. Even though the
mean weight is close to the target weight of 5.5 grams, the standard deviation of 0.106 indicates that
about 75% of the teabags will fall within 0.212 grams around the target weight of 5.5 grams. The
interquartile range of 0.13 also indicates that half of the teabags in the sample fall in an interval 0.13
grams around the median weight of 5.515 grams. The process can be adjusted to reduce the variation of
the weight around the target mean.
4. Problem 3.42
Excel output:
Kilowatt Hours
Mean
12999.22
Standard Error
546.9863
Median
13255
Mode
#N/A
Standard Deviation
3906.264
Sample Variance
15258895
Kurtosis
0.232784
Skewness
0.468115
Range
18171
Minimum
6396
Maximum
24567
Sum
662960
Count
51
(a)
mean = 12999.2158,
variance = 14959700.52,
std. dev. = 3867.7772
(b)
64.71%, 98.04% and 100% of these states have average per capita energy consumption within 1,
2 and 3 standard deviations of the mean, respectively.
(c)
This is consistent with the 68%, 95% and 99.7% according to the empirical rule.
(d)
Excel output:
Kilowatt Hours
Mean
12857.74
Standard Error
539.0489
Median
12999
Mode
#N/A
Standard Deviation
3811.651
Sample Variance
14528684
Kurtosis
0.464522
Skewness
0.494714
Range
18171
Minimum
6396
Maximum
24567
Sum
642887
Count
50
(d)
(a)
mean = 12857.7402, variance = 14238110.67, std. dev. = 3773.3421
(b)
66%, 98% and 100% of these states have average per capita energy consumption within
1, 2 and 3 standard deviation of the mean, respectively.
(c)
This is consistent with the 68%, 95% and 99.7% according to the empirical rule.
5. Problem 3.43
(a) Excel output (in \$ millions):
Pop Mean
Pop Std Dev
Market Cap
96822.61
70547.85603
(b) On average, the market capitalization for this population of 30 companies is \$96.8226 billions.
The average distances between the market capitalization and the mean market capitalization for this
population of 30 companies \$70.5479 billions.
```
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