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Stat 2470, Practice Exam #1, Spring 2014
Instructions: You may use your calculator for any functions the TI-84/83 model calculator is capable of
using, such as probability distributions and obtaining graphs of any data. To show work on these
problems, report the functions and their syntax as entered. Other things, such as integrating, must be
done by hand unless specifically directed otherwise. Round means to one more place than the original
data, variances and standard deviations to two more than the original data. Round probabilities to
three significant figures or use exact values. In order to receive partial credit on any problem, you must
show some work or I will have nothing to award partial credit on. Be sure to complete all the requested
parts of each problem.
1. A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the
accompanying data on time (sec) to complete the escape:
373
356
392
370
359
369
364
363
374
366
375
359
364
424
356
325
325
403
339
394
334
393
402
397
a.
Construct a stem-and-leaf display of the data. How does it suggest that the sample mean
and median will compare?
b.
Calculate the values of the sample mean and median.
c.
By how much could the largest time, currently 424, be increased without affecting the value
of the sample median? By how much could this value be decreased without affecting the
value of the sample mean?
d.
e.
What are the values of
and
when the observations are re-expressed in minutes?
Construct a boxplot of the data. Are there any outliers?
2. Calculate and interpret the values of the sample mean and sample standard deviation for the
following observations on fracture strength.
128
131
142
168
87
93
105
114
96
98
3. At a certain gas station, 40% of the customers use regular unleaded gas ( ), 35% use extra
unleaded gas ( ), and 25% use premium unleaded gas ( ). Of those customers using
regular gas, only 30% fill their tanks (event B). Of those customers using extra gas, 60% fill
their tanks, whereas of those using premium, 50% fill their tanks.
a. What is the probability that the next customer will request extra unleaded gas and fill the
tank?
b. What is the probability that the next customer fills the tank?
c. If the next customer fills the tank, what is the probability that regular gas is requested?
Extra gas? Premium gas?
4. Two pumps connected in parallel fail independently of one another on any given day. The
probability that only the older pump will fail is .15, and the probability that only the newer
pump will fail is .05. What is the probability that the pumping system will fail on any given day
(which happens if both pumps fail)?
5. A real estate agent is showing homes to a prospective buyer. There are ten homes in the
desired price range listed in the area. The buyer has time to visit only four of them.
a.
In how many ways could the four homes be chosen if the order of visiting is considered?
b.
In how many ways could the four homes be chosen if the order is disregarded?
c.
If four of the homes are new and six have previously been occupied and if the four
homes to visit are randomly chosen, what is the probability that all four are new? (The
same answer results regardless of whether order is considered.)
6. Suppose that vehicles taking a particular freeway exit can turn right (R), turn left (L), or go
straight (S). Consider observing the direction for each of three successive vehicles.
a.
List all outcomes in the event A that all three vehicles go in the same direction.
b.
List all outcomes in the event B that all three vehicles take different directions.
c.
List all outcomes in the event C that exactly two of the three vehicles turn right.
d.
List all outcomes in the even D that exactly tow vehicles go in the same direction.
e.
List outcomes in
,C
D, and C
D.
7. A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The
geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis.
a.
What is the pmf (pdf) of the number of granite specimens selected for analysis?
b.
What is the probability that all specimens of one of the two types of rock are selected
for analysis?
c.
What is the probability that the number of granite specimens selected for analysis is
within 1 standard deviation of its mean value?
8. Use the cumulative binomial probabilities table available in your text to obtain the following
probabilities:
a.
B(4; 10, .4)
b.
b(4; 10, .4)
c.
b(6; 10, .6)
d.
when
Bin(10,.4)
e.
when
Bin(10,.3)
f.
when
Bin(10,.7)
g.
when
Bin(10,.3)
9. Automobiles arrive at a vehicle equipment inspection station according to a Poisson process
with rate = 10 per hour. Suppose that with probability .5 an arriving vehicle will have no
equipment violations.
a.
What is the probability that exactly ten arrive during the hour and all ten have
no violations?
b.
For any fixed
what is the probability that y arrive during the hour, of
which ten have no violations?
c.
What is the probability that ten “no-violation” cars arrive during the next hour?
[Hint: Sum the probabilities in part (b) from y = 10 to
]
10. Let X denote the amount of space occupied by an article placed in a
The pdf of X is
packing container.
a.
Obtain the cdf of X.
b.
What is
c.
What is
d.
What is the 75th percentile of the distribution?
e.
Compute
f.
What is the probability that X is within 1 standard deviation of its mean value?
?
11. In each case, determine the value of the constant c that makes the probability statement
correct.
a.
b.
c.
d.
e.
12. Suppose that when a transistor of a certain type is subjected to an accelerated life test, the
lifetime X (in weeks) has a gamma distribution with mean of 40 and variance of 320.
a. What is the probability that a transistor will last between 1 and 40 weeks?
b. What is the probability that a transistor will last at most 40 weeks? Is the median of
the lifetime distribution less than 40? Why or why not?
Answers:
1. a.
A. stem-and-leaf display of this data appears below:
32
33
34
35
36
37
38
39
40
41
42
55
49
stem: ones
leaf: tenths
6699
34469
03345
9
2347
23
4
The display is reasonably symmetric, so the mean and median will be close.
b. The sample mean is = 9638/26 = 370.7. The sample median is = (369+370)/2=369.50.
c. The largest value (currently 424) could be increased by any amount. Doing so will not change the fact
that the middle two observations are 369 and 170, and hence, the median will not change. However,
the value x = 424 can not be changed to a number less than 370 ( a change of 424-370 = 54) since that
will lower the values(s) of the two middle observations.
d. Expressed in minutes, the mean is (370.7 sec)/(60 sec) = 6.18 min; the median 6.16 min.
2. The sample mean,
The sample standard deviation,
On average, we would expect a fracture strength of 116.2. In general, the size of a typical deviation from
the sample mean (116.2) is about 25.75. Some observations may deviate from 116.2 by more than this
and some by less.
3.
Therefore,
a. P
= .21
b. P(B) =
c. P(
4. Let = older pump fails,
x, and x = P(
) = P(
= newer pump fails, and x = P(
). The P( ) = .15 + x, P(
The resulting quadratic equation,
) = .05 +
has roots x = .0095 and x = .7905. Hopefully the smaller root is the actual
probability of system failure.
5. a.
b.
(10)(9)(8)(7) = 5040
=
c. P(all 4 are new) = (# of ways of visiting all new)/(# of ways of visiting) =
=
6. a. Event A = {RRR, LLL, SSS}
b. Event B = {RLS, RSL, LRS, LSR, SRL, SLR}
c. Event C = {RRL, RRS, RLR, RSE, LRR, SRR}
d. Event D = {RRL, RRS, RLR, RSE, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL, SRS, SLS, RSS,
LSS}
e. Event
contains outcomes where all cars go the same direction, or they all go different directions:
= {RRR, LLL, SSS, RLS, RSL, LRS, LSR, SRL, SLR}
Because Event D totally encloses Event C, the compound event C D = D. Therefore, C D = {RRL,
RRS, RLR, RSR, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL, SRS, SLS, RSS, LSS}
Using similar reasoning, we see that the compound event C D = C. Therefore, C D = {RRL, RRS,
RLR, RSR, LRR, SRR}
to be more than 10 of the basaltic).
7. Following the same pattern for the other values, we arrive at the pmf, in table form below.
x
P(x)
5
.0163
6
.1354
7
.3483
8
.3483
9
.1354
10
.0163
b. P(all 10 of one kind or the other) = P(X = 5) + P(X = 10) = .0163 + .0163 + .0326
c. E(X) = n . ( M / N) = 15 (10 / 20) = 7.5;
V(X) = (5/19) (7.5) [1 – (10/20)] = .9869; = .9934
7.5 .9934 = (6.5066, 8.4934), so we want
P( X = 7) + P( X = 8) = .3483 + .3483 = .6966
8. a.
B(4;10,.4) = .633
b. b(4;10,.4) = B(4;10,.4) – B(3;10,.4) = .251
c. b(6;10,.6) = B(6;10,.6) – B(5;10,.6) = .251
d.
e.
f.
g.
9. a.
= (.000977)(.125) =.000122
b. P(y arrive and exactly 10 have no violations)
= P(exactly 10 have no violations / y arrive). P(y arrive)
c. P(exactly 10 without a violation) =
In fact, generalizing this argument shows that the number of “no-violation” arrivals within the
hour has a Poisson distribution with parameter 5; the 5 results from
10. a.
Therefore,
b.
c.
d. The 75th percentile is the value of x for which F(x) = .75
e.
f.
=.8465 - .1602 = .6863
11. a.
b.
c.
d.
e.
12.
a.
b.
so while the mean is 40, the median is less than 40.
This is the result of the positive skew of the gamma distribution.