Homework Help 1234
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quiz1.pdf
STAT 200 QUIZ 1 Solutions Section 6380 Fall 2013
• The quiz covers Chapters 1, 2 and 3.
(For Questions 1 to 6) The checkout times (in minutes) for 12 randomly selected customers at a large supermarket during the store’s busiest time are as follows:
4.6 8.5 6.1 7.8 10.7 9.3
12.4 5.8 9.7 8.8 6.7 13.2
1. (2 points) (Show work) What is the mean checkout times based on the 12 checkout times.
The mean checkout time based on the raw data
= (4.6 + 8.5 + 6.1 + 7.8 + 10.7+ 9.3 + 12.4 + 5.8 + 9.7 + 8.8 + 6.7 + 13.2) / 12 = 103.6 / 12
= 8.63
2. (6 points) (Show work) Give a 5-number summary of the checkout time, and construct the corresponding boxplot.
Arrange the checkout times from low to high:
Method 1 : Use the approach from the book
Location 1 2 3 4 5 6 7 8 9 10 11 12 Data 4.6 5.8 6.1 6.7 7.8 8.5 8.8 9.3 9.7 10.7 12.4 13.2
Minimum: 4.6
Q1: Find the location L of the 25th percentile in the sorted list:
25 112 .12 3 100 100 4
kL n= ⋅ = ⋅ = =
Because L = 3 is a whole number, the 25th percentile (first quartile) is located between 3rd and 4th checkout time in the sorted list. The 3rd and 4th checkout times are found to be 6.1 and 6.7, so the first quartile Q1 is (6.1 + 6.7) / 2 = 6.4 minutes
Median: Find the location L of the 50th percentile in the sorted list:
50 112 .12 6 100 100 2
kL n= ⋅ = ⋅ = =
Because L = 6 is a whole number, the 50th percentile (median) is located between 6th and 7th checkout time in the sorted list. The 6th and 7th checkout times are found to be 8.5 and 8.8, so the median is (8.5 + 8.8) / 2 = 8.65 minutes
Q3: Find the location L of the 75th percentile in the sorted list:
75 312 .12 9 100 100 4
kL n= ⋅ = ⋅ = =
Because L = 9 is a whole number, the 75th percentile (third quartile) is located between 9th and 10th checkout time in the sorted list. The 9th and 10th checkout times are found to be 9.7 and 10.7, so the third quartile Q3 is (9.7 + 10.7) / 2 = 10.2 minutes
Maximum: 13.2
Arrange the checkout times from low to high:
Method 2: Use the MATH106 approach
Location 1 2 3 4 5 6 7 8 9 10 11 12
Data 4.6 5.8 6.1 6.7 7.8 8.5 8.8 9.3 9.7 10.7 12.4 13.2
10.7 12.4 13.2
Minimum: 4.6
First quartile Q1 is the average of 3rd and 4th number: (6.1 + 6.7) / 2 = 6.4 minutes
Median is the average of the 6th and 7th number : (8.5 + 8.8) / 2 = 8.65 minutes
Third quartile Q3 is the average of 8th and 9th number: (9.7 + 10.7) / 2 = 10.2 minutes
Maximum: 13.2
3. (4 points) Prepare a frequency distribution with a class width of 2 minutes (with the first class as 4.0 – 5.9, etc.).
Checkout Time Frequency 4.0 - 5.9 2 6.0 - 7.9 3 8.0 - 9.9 4 10.0 - 11.9 1
12.0 - 13.9 2
4. (4 points) Based on the frequency distribution from Question 3, construct a histogram.
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
4.95 6.95 8.95 10.95 12.95
Frequency
Checkout Time (Minutes)
5. (4 points) (Show work) We usually find the mean checkout time based on the raw data. Now suppose that we don’t have the raw data, but instead, we only have the frequency distribution. Find the mean checkout time based on the frequency distribution from Question 3. Compare your results with the "actual" mean checkout time you found in Question 1.
midpoint (x) frequency (f) x * f
4.95 2 9.9 6.95 3 20.85 8.95 4 35.8
10.95 1 10.95
12.95 2 25.90
Total 12 103.4
Mean = 103.4 / 12 = 8.62 minutes
The mean checkout time based on the frequency distribution is not equal to the “actual” mean (8.63), but it’s a good approximation of the actual mean.
6. (4 points) (Show work) Find the standard deviation of the checkout time based on the frequency distribution from Question 3.
x f x2 f * x2 f * x
4.95 2 24.5025 49.005 9.9
6.95 3 48.3025 144.9075 20.85
8.95 4 80.1025 320.41 35.8
10.95 1 119.9025 119.9025 10.95
12.95 2 167.7025 335.405 25.9
sum 12
969.63 103.4
2 2 2[ ( )] [ ( )] 12(969.63) (103.4) 944 7.1515 2.67
( 1) (12)(11) 132
n f x f x s
n n
− −
= = = = = −
∑ ∑
_____________________________________________________________________
(For Questions 7, 8 & 9) To qualify for a mortgage, an applicant needs to have a good FICO credit score. We chose a random sample of 6 applicants from a local bank, and their FICO scores are 714, 751, 664, 820, 550 and 613.
7. (5 points) (Show work) What is the standard deviation of the FICO scores in the sample?
x
Method 1 : Use Formula 3-5
x2 714 509796 751 564001 664 440896 820 672400 550 302500 613 375769
∑x =4112 ∑x2 = 2865362
The sample variance
2 2 2 2 ( ) ( ) 6(2865362) (4112) 9454.27
( 1) (6)(5)
9454.27 97.23
n x x s
n n
s
− − = = =
−
= =
∑ ∑
Method 2: Use Formula 3-4
x x-mean (x-mean)2 714 28.67 821.9689 751 65.67 4312.5489 664 -21.33 454.9689 820 134.67 18136.0089 550 -135.33 18314.2089 613 -72.33 5231.6289
∑ =4112 ∑ = 47271.3334
2 2
4112 685.33 6
( ) 47271.3334 9454.27, 1 5
9454.27 97.23
x x
n x x
s n
s
= = =
− = = =
− = =
∑
∑
8. (2 points) (Show work) Are any of these FICO scores considered unusual in the sense of our textbook? Explain.
minimum “usual” value = 685.33 – 2( 97.23) = 490.87
maximum “usual” value = 685.33 + 2(97.23) = 879.79
The given FICO scores fall between 490.87 and 879.79, so none of the FICO scores is considered unusual.
9. (2 points) (Show work) What is the coefficient of variation in FICO scores of the sample?
4112 685.33 6
x = =
97.23100% 100% 14.19% 685.33
sCV x
= = =
____________________________________________________________________________
10. (4 points) Determine whether the given value is a parameter or statistic. Please explain your answer.
(a) If an agency conducts a survey on the choice of vegetable by Americans. It is reported by the agency that 57% of Americans prefers broccoli.
This reported percentage is a statistic since survey does not include the whole American population; it only conducted on a sample.
(b)A UMUC STAT 200 instructor randomly selected 100 students and found the average study time is 20.5 hours per week.
The result was based on a random sample, so the reported average is a statistic.
___________________________________________________________________________
11. (3 points) Mimi is a recent college graduate, who is considering some career opportunities. One factor that will affect her decision is how much money she is likely to make. She got salary information from Internet. Since you are taking a Statistics course, she would like to get your opinion on whether mean salary or median salary information is more useful for her. What would be your recommendation? Please explain your answer. Median salary is more useful for her since median can be thought of loosely as a “middle value” in the sense that about half of the salaries are below the median and half are above it. In addition, unlike the mean, the median is not impacted by the extreme values.
