Download Spring 2001 Test 2

These problems are meant to help you study. The presence of a
problem on this sheet does not imply that there will be a similar
problem on the test. And the absence of a problem from this sheet
does not imply that the test does not have such a problem.
----The game called “rock-paper-scissors” involves players selecting one of those three options (rock or paper or
scissors). I recently read that novice players tend to pick “scissors” less often than would be expected if the three
choices were equally likely.
Identify (in words) the relevant parameter in this study.
The parameter is the actual proportion of the time that a novice player picks “scissors.”
The articles that I read described a study in which 119 novice players played this game against a computer, and
researchers recorded the first option that each player selected.
Under equally likely the expected number of scissors is? 39.67
The data resulted in fourteen (14) of the 119 players selecting “scissors” as their first option.
Describe how you could use a fair six-sided die to conduct a simulation analysis of this study and its result. Give
sufficient detail that someone else could implement this simulation analysis based on your description. Be sure
to indicate how you would decide whether the observed data provide convincing evidence in support of the
claim that novice players tend to pick “scissors” less often than would be expected if the three choices were
equally likely.
Let the numbers 1 and 2 represent selecting “scissors,” so this has a 1/3 probability. Roll the die 119 times
(once to represent each player) and count the number of 1s and 2s. Then repeat this a larger number (say
1000) of times, counting the number of 1s and 2s in each set of 119 rolls. Then count how many of the
1000 repetitions produced 14 or fewer for the number of 1s and 2s, and divide by the number of
repetitions to produce an approximate probability-value. If this approximate probability-value is very
small (say, less than 0.05), then conclude that the data provide strong evidence that scissors is really
selected less than 1/3 of the time.
6. The number of court cases won by the a local law firm during any given week has the following probability
distribution:
x
0
1
2
3
p(x)
?
0.50
0.25
0.15
i) What is the probability that no court case is won? [[0.10]]
ii) What is the probability that at least one court case is won? [[0.90]]
iii) What is the mean number of court cases won? [[1.45]]
8. At the MU bookstore, 30% of students need assistance finding their books to purchase at the
beginning of any given semester. A random sample of 10 students is selected.
i) What is the value of n? [[10]] and p? [[0.3]]
ii) P(Exactly 4 need assistance) = [[0.20012]]
iii) P(More than 6 need assistance) = [[0.01059]]
iv) What is the average number expected to need assistance? [[3]]
x
0
1
2
3
4
5
6
7
8
9
10
P(X <= x)
0.02825
0.14931
0.38278
0.64961
0.84973
0.95265
0.98941
0.99841
0.99986
0.99999
1.00000
9. Which of the following will have a Binomial Distribution: [[B]]
A) The number of red cards out of 10 randomly selected cards without replacement from a well-shuffled deck.
B) The number of times a die comes up even in 15 throws.
C) The number of parking tickets that the U of A gives on any given week.
D) The number of calls received by a call-center in any given hour.
11. Find the percentage from the standard normal tables for the following:
[4 pt each]
a) below z = 2.05 ____A______
(A) 97.98
(B) 99.38
(C) 2.02
(D) 47.48
b) between z = 0 and z = 3.00 _____D_____
(A) 99.87
(B) 50
(C) 149.87
(D) 49.87
c) above z = –1.00 _____B_____
(A) 34.52
(B) 84.13
(C)15.87
(D) none
A manufacturer says that not only will 90% of their copiers last at least 36 months, 61% will last at least 42 months.
If the copier lifetimes are normally distributed, what normal model parameters is this manufacturer claiming?
[[ 36 is the 10th percentile 1.28 stdev below the mean
[[ 42 is the 39th percentile 0.28 stdev below the mean
[[ 36 = mean – 1.28 stdev
[[ 42 = mean – 0.28 stdev
[[ solve mean and stdev
[[ mean = 43.68 and stdev = 6
12. The A.C.Nielsen Company reported that the mean weekly television viewing time for children aged 2-11 years is 25 hours.
Assume that the weekly television viewing times of such children are normally distributed with a standard deviation of 5 hours.
a) What percentage of such children watched television for less than 33 hours?
(Mark and label the information given on the adjoining diagram)
[5]
94.52 %
b) If a child is in the highest 15% of the children who watch television, what is the least number of hours he/she must be
watching television per week?
[7]
30.2 hours
13. If your z-score on this test is 2.00, which would mean that…__B_______
A) you scored 2 points more than the average score of the class
B) your score is 2 standard deviations more than the average score of the class
C) your score is 2 times the average score of the class
D) your score is 2 times the standard deviation of the class
The test scores on a placement exam are normally distributed random variables
with a mean of 104 and standard deviation of 12.
12) Find the probability that a selected test score is less than 117 [[0.8599]]
13) Find the probability that a selected test score is between 92 and 122 [[0.7745]]
14) Find the placement test score that has 75% of the other scores below it. [[104 + 0.67(12) = 112.04]]
The times to complete a placement exam have a mean of 40 minutes and a standard deviation of 8
minutes. The exam is given to 49 students.
15) The sampling distribution of the sample mean scores has a mean given by _____. [[40]]
16) The sampling distribution of the sample mean has a standard deviation given by _____. [[1.143]]
17) What is the probability that the sample mean will be greater than 42 minutes ______. [[0.0401]]
  E ( X )   x P( x)
binomial P( x) 

 (x  )
2
P( x)
n!
( p) x (1  p) n  x binomial   E ( X )  np binomial   np(1  p)
x! (n  x)!
x
z
x    z

x  
x 

n
Z 
x  x
x
9-15 [25 pts]
A county council is made up of 100 citizens chosen at random to represent the entire
population of the county. 37% of the citizens in this county are minorities.
Binomial with
n = 100
p = 0.37
9. Let X be the number of minority citizens who are selected for the county council.
Is X a Discrete Random Variable or a Continuous Random Variable? Justify your answer.
Discrete RV – possible values = 0,1,2,…,100 LIST  Binomial n=100 p=0.37
10. How many minority citizens would be expected to be selected for the county council?
np = 100(0.37) = 37
11. Certainly, we cannot guarantee that the expected value will be observed, so calculate the
standard deviation of the number of minority citizens selected for the county council.
sqrt[np(1-p)] = sqrt[100(0.37)(0.63)] = 4.83
12. Using the table to the right, find the probability that between 25 and 40, inclusive,
county council members will be minorities?
P[X <= 40] – P[X <= 24] = 0.76698 – 0.00392 = 0.76306
13. Using the table to the right,
what is the probability that exactly 37 county council members will be minorities?
P[X <= 37] – P[X <= 36] = 0.54474 – 0.46234 = 0.0824
14. Using the table to the right,
what is the probability that more than half of the county council members will be minorities?
P[X > 50] = 1 – P[X <= 50] = 1- 0.99705 = 0.00295
15. When the members of the county council are selected, 60 members turn out to be
minority citizens. The Election Caucus claims that this county council is biased and could not
have been selected randomly. Do you think the Election Caucus is reasonable in their
complaint? Explain.
Yes, expect 37 minorities!
From the graph  60 is very unlikely.
Binomial, n=100, p=0.37
0.09
0.08
Random selection is suspect …
0.07
Probability
60 is almost 5 stdev above the mean!!
0.06
0.05
0.04
0.03
0.02
0.01
0.00
0
20
40
60
X
80
100
x
0
1
…
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
…
99
100
P(X<=x)
0.00000
0.00000
…
0.00000
0.00001
0.00003
0.00007
0.00018
0.00043
0.00095
0.00198
0.00392
0.00739
0.01327
0.02273
0.03722
0.05835
0.08771
0.12666
0.17597
0.23565
0.30473
0.38122
0.46234
0.54474
0.62498
0.69989
0.76698
0.82465
0.87223
0.90991
0.93859
0.95954
0.97426
0.98419
0.99063
0.99464
0.99705
0.99843
0.99920
0.99960
0.99981
0.99991
0.99996
0.99998
0.99999
1.00000
1.00000
1.00000
…
1.00000
1.00000
This may be helpful ...
Finding areas associated with normal curves:
1.
2.
3.
4.
Draw a picture showing the Y values and desired area.
Switch from natural units to std. dev. units (Z-scores)
Refer to the Z table,T.I.,minitab,technology to obtain area.
Do the appropriate arithmetic to get the final area you need.
Note:
The Z distribution is the "Standard Normal Distribution".
It has a mean of 0 and a standard deviation of 1.0.
Here is an approach for backward problems:
1.
2.
3.
Draw a picture showing the given areas shaded.
Look up the appropriate area in Z table,T.I.,minitab,technology
and find the z that matches.
(That is -- use the Z table backwards -- invCDF.)
Change back to original units: doing necessary arithmetic