MATH 130 Instructor: KSR - GOOD LUCK TO: __________________________________________________
EXAM 1 – SPRING 2016
Instructions:
- You may use a hand-calculator for this test, but not a cell phone or laptop or … .
- You may use a one-sided “cheat” sheet.
- All steps must be clearly shown for computational problems.
- No scratch paper is permitted.
__a__ 1. [2 pts] A study of Millersville University students recorded the age of each respondent …
and if they were a commuter [yes or no] for each respondent. Select the best answer:
(a)
(b)
(c)
(d)
age is a quantitative variable and commuter is a categorical variable.
age and commuter are both categorical variables.
age and commuter are both quantitative variables.
age is a categorical variable and commuter is a quantitative variable.
2. Create a dataset of 5 numbers …
[5 pts]
… such that the median and mode are each smaller than 20 but the mean is larger than 40.
…lots of correct answers … 0,0,0,0,999
3. [12 pts] The weight of an organ in adult males has a bell-shaped distribution with …
… a mean of 300 grams and a standard deviation of 30 grams.
3.a) About 95% of organs will be between what weights? __240__ and __360__
68-95-ALL RULE
300-2(30), 300+2(30)  mean – 2(stdev), mean + 2(stdev) gives 95%
3.b) the percentage of organs weighing between 270 grams and 330 grams is approximately …__68%__
68-95-ALL RULE
300-1(30), 300+1(30)  mean - 1stdev, mean + 1stdev gives 68%
3.c) the percentage of organs weighing between 240 grams and 330 grams is approximately …__81.5%__
68-95-ALL RULE
300-2(30), 300+1(30)  mean - 2stdev, mean + 1stdev gives 47.5% + 34% = 81.5%
__b__ 4. [2 pts]
To display the distribution of religion {Christian, Jewish, Muslim, …] among students in a course,
it would be correct to use
(a) a stem-leaf but not a bar chart.
(c) neither a stem-leaf nor a bar chart.
(b) a bar chart but not a stem-leaf.
(d) either a stem-leaf or a bar chart.
1
5. [10 pts] The following graph displays the final exam performance by format for an instructor who taught a
course in 3 formats: 2Day, 4Day, Evening.
Boxplot of 2Day, 4Day, Eve
Based on these graphs,
what do you suspect is the shape of all three distributions?
2Day
Left-Tailed – Left Skew
4Day
Based on these graphs,
which format tends to have the worst performance?
Eve
Evening
3 0 3 5 40 4 5 5 0 5 5 6 0 65 7 0 7 5 8 0
Based on these graphs,
which format has the least variability in performance?
85 9 0 9 5 1 00
Data
4-Day
Based on these graphs, which format had the largest maximum? Give Its Value.
4-Day -- 95
Based on these graphs, which format had the smallest Q1, 25th percentile? Give Its Value.
Evening -- 50
6. Match the histograms to the numerical summary statistics … [8 pts]
EXAM
#1
#4
#2
#3
MEAN
75
75
75
75
MEDIAN
75
75
75
75
STDEV
1.777
22.33
10.91
15.69
Histogram of exam1, exam2, exam3, exam4
50
60
70
80
90
100 50
60
exam1
70
80
90
100
exam2
100
75
Percent
50
25
exam3
100
0
exam4
75
50
25
0
50
60
70
80
90
100 50
60
70
80
90
100
2
7. [27 pts] Consider the two-way table
giving updated info for the classification
of 1,000 USA adults classified by marital
status and educational level attained.
Less than HS
Been Married
Never Married
What is the probability that … (5.a-5.f)
65
60
125
HS
Some College
Degree College Degree
128
85
246
524
174
83
159
476
302
168
405
1000
7.a) … a person has never married? 476/1000 = 47.6%
7.b) … a person has college degree? 405/1000 = 40.5%
7.c) … a person does not have college degree? 595/1000 = 59.5%
7.d) … a person has never married and has college degree? 159/1000 = 15.9%
7.e) … a person has never married or has college degree? 722/1000 = 72.2% = 47.6% + 40.5% - 15.9%
7.f) … a person has never married, given that the person has college degree?
159/405 = 15.9%/40.5% = 39.259%
7.g) Are the events “never married” and “college degree” mutually exclusive events? Why?
NO – the intersection exists – 159/1000 --- “Never Married” and “College Degree”.
7.h) Are the events “never married” and “college degree” complementary events? Why?
NO –
the complement of “Never Married” is “Been Married” …
the complement of “College Degree is “not College Degree” …
7.i) Are the events “never married” and “college degree” independent events? Why?
NO – since
P(Never Married) = 476/1000 = 47.6% (a)
not equal
P(Never Married | College Degree) = 159/405 = 39.259% (f)
8. Millersville University in order to investigate the number of courses students are currently enrolled in
conducted a survey resulting in the following data for a sample of 3 students: 2, 5, 2
Quantify the variation by calculating the range & standard deviation for this dataset.
{8 pts} range = 5-2 = 3
2
2
5
2-3 = -1
2-3 = -1
5-3 = +2
1
1
4
6/[3-1]= 3
sqrt(3) = 1.732… = stdev
mean = 9/3 = 3
n = 3
3
9. {4 pts} Determine the type of sampling (simple random, stratified, cluster, systematic) for each:
(9.a) A manager at Apple selects every 12th iPod off the assembly starting with the ninth. __systematic__
(9.b) Jill divides her day into three parts: morning, afternoon, and evening.
She then measures her mood at 4 randomly selected times during each part of the day. __stratified____
10. A church softball league has 15 teams; however, because of fights they have decided to … {8 pts}
(10.a) just randomly select 4 teams to make the playoffs – how many ways are there to select these 4 teams?
{(15)(14)(13)(12)}/{(4)(3)(2)(1)} = 15C4 = 1,365
(10.b) On second thought, they just randomly select the 1st, 2nd, 3rd, and 4th place finishers from the 15 teams
…-- how many ways are there to select these four finishers?
(15)(14)(13)(12) = 15P4 = 32,760
11. {8 pts}
From the survey-data, 78% of students had dog preference, and 58% of students were coffee drinkers.
In fact, 47% of students had dog preference AND drink coffee.
Suppose a student is randomly selected …
[Hint: Draw Table!]
Dog
Cat
Coffee Yes
0.47
0.11
0.58
Coffee No
0.31
0.11
0.41
0.78
0.22
1
What is the probability that they have a cat preference AND drink coffee? 0.11
What is the probability that they have a cat preference OR drink coffee? 0.69
12. [9 pts] About 22% of the population of a large country is allergic to ragweed.
If two people are randomly selected … Find the probability that … [Hint: Construct Tree!]
both people will be allergic to ragweed. (0.22)(0.22) = 0.0484
neither of the people will be allergic to ragweed. (0.78)(0.78) = 0.6084
at least one person will be allergic to ragweed. 1 – 0.6084 = 0.3916
4