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Summary-line: 23-Apr PBleaks@aol.com [178] #Mat 108 extra credit
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4-22-98
Professor Rodriquez,
My name is Carrie Bleakley 126-70-1480, I am in your MAT 108 class at
2:30 on Tuesdays and thursday. He are the ten questions you told me I could
e-mail to you for extra credit. Thank you.
Probability Problems
Question 1) There are 60 cars in a parking lot: 14 Chevrolets, 12 Fords, 6
Plymouths, 5 Buicks, 4 Pontiacs and 19 Foreign cars. A women crosses the
street and enters the parking lot to retrieve her car. What is the
probability that she is the owner of a Ford?
a) .80
b) .32
c) .20
d) .68
e) .16
Answer 1) probability = # of ways the event in question can occur / # of
events
possible
p = 12/60 = .20 choice c is correct.
Question 2) Using the information from Question 1, what is the probability
that she owns a Foreign car?
a) .80
b) .20
c) .16
d) .32
e) .68
Answer 2) p= # of ways the event in questin can occur / # of events possible
p = 19/60 = .32 Choice d is correct.
Question 3) A manufacturer of hockey equipment is designing a new set of
protective gear for college hockey players, and the design engineers are
basing their model on an average college male chest measurement of 39 inches
with a standard deviation of 2.5 inches. We will assume that chest
circumferences are normally distributed in the college male population. What
percentage of the population has a chest measurement of 44 inches or so?
a) 47.72%
b) 23.86%
c) 4.56%
d) 2.28%
e) not enough information
Answer 3) After drawing the curve, we calculate the z score for 4, which is
z=(44-39)/2.5 = 2.0. under the normal table the area is
95.45 from
-2.0 to 2.0, therefore (100-95.45)/ 2 = 2.28% with a chest
measurement
of 44 inches or so. Choice d is correct.
Question 4) using the information in Question 3, What is the probability that
a hockey player, chosen at random, would have a chest measurement of less than
37 inches?
a) .21
b) .28
c) .76
d) .10
e) .79
Answer 4) After drawing the curve, we calculate the z score for 37, which is
z= (37-39)/2.5 = -0.8. Under the normal curve 57.63 is the
area between
-0.8 and 0.8, therefore (100-57.63)/2 = 21.19% below this point,
so the
probability that any hockey player chosen at random would have
a chest
measurement of less than 37 inches would be p=.21. Choice a.
Question 5) According to the National Center for Health Statistics the
average male college student weighs 154 lbs. and the distribution of weights
has a standard deviation of 21.4 lbs.. Assume that this distribution is
normally distributed in the population to answer the FIVE questions. What
percent of college men weigh 200 lbs. or more?
a) 2.149%
b) .60%
c) 1.58%
d) 4.298%
e) .967%
Answer 5) You need to convert into standard units, (200-154)/2.4 = 2.15
look in table the area is 96.84 from -2.15 to 2.15, then
(100-96.84)/2 = 1.58% will weigh 200 lbs. or more, Choice c.
Question 6) Tami tells her friends that she will not date any man that weighs
less than she does. Tami weighs 125. What percent of the college men need
not bother to call?
a) 12.86%
b) 19.36%
c) 4.84%
d) 9.68%
e) 90.32%
Answer 6) After drawing the curve, we calculate the z score for 125, which is
z= (125-154)/21.4 = -1.3, look at table, the areafrom -1.3 to
1.3 is
80.64, so (100-80.64)/2 = 9.68% of the college men need not
bother
to call. Choice d.
Question 7) Bill states that he is heavier than 99% of the college male
population. How much would Bill have to way to make this statement?
a) 218.2 lbs.
b) 196.8 lbs
c) 201 lbs.
d) 209.64 lbs.
e) not enough information
Answer 7) You look at the table and find the area closest to 99, which is
99.07; so
the z is 2.60. Then take the z and multiply it times the sd
and add that
value to the average.
2.60(21.4)+154= 209.64 lbs. Choice d.
Question 8) A hot air ballonist wants to hire a male assistant who weighs 140
lbs. or less. What percent of the college male population would be eligible
for this job?
a) 51.57%
b) 12.89%
c) 48.43%
d) 17.57%
e) 25.79%
Answer 8) We calculate the z score for 140, which is z=(140-154)/21.4 = -.65
In the table the area from -.65 to .65 is 48.43, therefore
(100-48.43)/2 =
25.79% would be eligible for the job.
Question 9) The middle 99% of the college male population would be included
between what two weights?
a) 94.52 to 203.86
b) 98.36 to 209.64
c) 99.7 to 199.7
d) 132.6 to 175.4
e) 111.2 to 196.8
Answer 9) We use the table to find the closest area to 99, we then get a z of
2.60,
then -2.60(21.4)+154= 98.36 lbs and 2.60(21.4)+154= 209.64
lbs.
The middle 99% would be between, 98.36 lbs. and 209.64 lbs.
Question 10) What weight corresponds to the 90th percentile?
a) 164.7 lbs.
b) 207.5 lbs
c) 189.31 lbs.
d) 168.98 lbs.
e) 175.83 lbs.
Answer 10) Using the table we find the closest area to 40% ( the table gives
area
symmetrical around z), z= .50, therefore .5 (21.4) +154 =
164.7 lbs.
Choice a.