Download practice exercises for exam 2

THE UNIVERSITY OF NEVADA RENO
DEPARTMENT OF MATHEMATICS AND STATISTICS
STAT 152, FALL 2008
PRACTICE EXERCISES FOR EXAM 2
INSTRUCTOR: Dr. Tomasz J. Kozubowski
INSTRUCTIONS
The following problems are a good practice for EXAM 2. Your homework problems are also great practice. The
problems (and their number) on the exam however, may be different from those on this practice sheet or
homeworks. In particular, exam problems will all be multiple choice. Additionally, exam problems will cover
all material in chapters 5 - 7.
On the exam, you will see the following instructions. Please make sure you know all the info requested.
Each problem is worth 1 point. You must shade your answers well on the scantron sheet. In order for your exam
to be graded, you must write your name in BLOCK CAPITAL LETTERS, LAST NAME FIRST in the NAME
field, your R number in the SUBJECT field, VERSION of the EXAM (A or B) in the TEST NO. field, TA
NAME in DATE field, and your section number in the PERIOD field.
You may use a calculator. No phones, pda's laptop's or other electronic devises are allowed. Your phones must
be switched off until you leave the room after the exam.
ENJOY!
1. It is known that 60% of adults favor capital punishment. Find the probability that on a jury of 12
members:
a) There will be a tie; that is six members will favor capital punishment;
(a) 0.1766
(b) 0.101
(c) 0.823
(d) 0.334
b) At most 2 will favor capital punishment.
(a) 0.9012
(b) 0.9912
(c) 0.0025
(d) 0.0028
2. In a study of long distance runners, the average weight was found to be 140 lbs. with a standard
deviation of 10 lbs. Assume the distribution of runners’ weights to be normal. Find the probability that
a randomly selected long distance runner weights:
(a) More than 150 lbs.
(a) 0.1587
(b) 0.0228
(c) 0.8413
(d) 0.9772
(b) At most 135 lbs.
(a) 0.3085
(b) 0.6915
(c) 0.9772
(d) 0.0228
(c) 0.8502
(d) 0.1498
(c) between 130 and 135 lbs.
(a) 0.6915
(b) 0.3085
1
3. Suppose there is a 80% chance that an insect egg hatches and becomes an adult. If a “mommy
insect” lays 100 eggs, what is the probability that:
(a) at least 60 will become adults
(a) 1
(b) 0
(c) 0.5
(d) 0.1498
(b) less than 70 will become adults
(a) 0.6915
(b) 0.5
(c) 0.0062
(d) 0.9938
4. The heights of adult men follow a normal distribution with mean 1.8 meters and standard deviation
of 0.1 meters. Only the tallest 10% of men can get to the “TALL MEN CLUB”. What is the minimal
height of a man who can get into that club?
(a) 1.928
(b) 1.929
(c) 3.08
(d) 2.02
5. A breeder of rabbits claims that he can breed rabbits yielding a mean weight of greater than 58
ounces. A random sample of 16 rabbits had a mean weight of 59.2 ounces and standard deviation of
weights s = 3 ounces. Assume normal distribution of rabbits’ weights. Find a 95% confidence interval
for the true mean weight of a rabbit.
6. A brewery distributes beer in cans labeled 12 oz. The Bureau of Weights and Measures randomly
selects 36 cans, measures their mean contents, and finds their mean weight to be 11.82 oz. Assume
that  is known to be 0.38 oz. Find a 95% confidence interval for the true mean content of a can.
7. In the context of problem 6, how large a sample is required in order to be 95% confident that the
sample mean will not differ from the true mean content by more than 0.01.
(THIS COULD BE A MULTIPLE CHOICE QUESTION ON THE ACTUAL FINAL EXAM)
8. In their annual report, the State Department of Education stated that 62% of academic employees
in the State were permanent faculty and the remaining 38% were temporary. A professor in
Bumbleton State University (BSU) claims that BSU has a lower proportion of permanent faculty than
the overall State proportion of 62%. A random sample of 400 academic employees at BSU showed
215 of them to have permanent positions. Find a 90% confidence interval for the true proportion of
permanent academic employees at BSU.
9. In the context of Problem 8, how large a sample is required in order to be 99% confident that the
sample proportion will not differ from the true proportion of permanent faculty at BSU by more than
0.02. Use the sample information stated in the problem for your initial estimate of p.
(THIS COULD BE A MULTIPLE CHOICE QUESTION ON THE ACTUAL FINAL EXAM)
10. Assume that no prior information about the proportion of permanent faculty at BSU is available.
How large a sample is required to be 99% confident that the sample proportion of permanent faculty
at BSU will not differ from the true proportion by more than 0.02.
(THIS COULD BE A MULTIPLE CHOICE QUESTION ON THE ACTUAL FINAL EXAM)
11. Suppose that the weight of golf balls is normally distributed with mean of 9.5 grams and standard
deviation of 1.2 grams. I pack 36 golf balls into a bag. Find the probability that the bag will weigh less
than 360 grams? (Hint: Use the Central Limit Theorem)
(a) 0.9938
(b) 0.0062
(c) 0.3372
(d) 0.6628
2
KEY TO PRACTICE MIDTERM
1.
a) (a)
b) (d)
2.
a) (a)
b) (a)
c) (d)
3.
a) (a)
b) (c )
4. (a)
5. A breeder of rabbits claims that he can breed rabbits yielding a mean weight of greater than 58
ounces. A random sample of 16 rabbits had a mean weight of 59.2 ounces and standard deviation of
weights s = 3 ounces. Assume normal distribution of rabbits’ weights. Find a 95% confidence interval
for the true mean weight of a rabbit.
Let  =mean weight of rabbit. Then 95% confidence interval for  is
[ x - t (n-1) ,/2 . (s/ n ) , x + t (n-1) ,/2 . (s/ n )
]
= [ 59.2 – t (15,.025).(3/4) , 59.2 + t (15,.025).(3/4) ]
= [ 59.2 –2.131*.75 , 59.2 +2.131*.75 ]
= [ 57.60 , 60.79 ]
6. A brewery distributes beer in cans labeled 12 oz. The Bureau of Weights and Measures randomly
selects 36 cans, measures their mean contents, and finds their mean weight to be 11.82 oz. Assume
that  is known to be 0.38 oz. Find a 95% confidence interval for the true mean content of a can.
( X  Z / 2 

n
)  (11.82  (1.96)(0.38
36
))  (11.696, 11.944)
Hence, the true mean in between 11.696 and 11.944 with 95% confidence.
7. Refer to problem 6 above about the brewery. How large a sample is required in order to be 95%
confident that the sample mean will not differ from the true mean content by more than 0.01.
The correct answer is:
n = (z(alpha/2)*(sigma/m))^2
= ((-1.96)(0.38)/(0.01))^2
= 5547.2704
Hence, we would need a sample size of n = 5548 to be 95% confident that the sample mean will not
differ from the true mean content by more than 0.01 (margin of error).
8. In their annual report, the State Department of Education stated that 62% of academic employees
in the State were permanent faculty and the remaining 38% were temporary. A professor in
Bumbleton State University (BSU) claims that BSU has a lower proportion of permanent faculty than
the overall State proportion of 62%. A random sample of 400 academic employees at BSU showed
3
215 of them to have permanent positions.Find a 90% confidence interval for the true proportion of
permanent academic employees at BSU.
ANS.
(0.496491 , 0.578509)
9.How large a sample is required in order to be 99% confident that the sample proportion will not
differ from the true proportion of permanent faculty at BSU by more than 0.02. Use the sample
information stated in the problem for your initial estimate of p.
ANS.
n = 4125
10. Assume that no prior information about the proportion of permanent faculty at BSU is available.
How large a sample is required to be 99% confident that the sample proportion of permanent faculty
at BSU will not differ from the true proportion by more than 0.02.
Given m = 0.02,  = 1 – C = 1 – 99% = 0.01. Since no prior information for p0, we use p0 = ½.
n
Z / 22 p0 (1  p0 ) 2.5762 (0.5)(1  0.5)

 4147.36  4148
m2
0.022
11. Suppose that the weight of golf balls is normally distributed with mean of 9.5 grams and standard
deviation of 1.2 grams. I pack 36 golf balls into a bag. Find the probability that the bag will weigh
less than 360 grams. (Hint: Use the Central Limit Theorem.)
The Central Limit Theorem states that the sample mean of any set of observations has approximately
a normal distribution if n is sufficiently large. So the expected value of the mean of X is 9.5 grams
and the standard deviation of the mean of X is (1.2)/ 36) = (1.2)/6 = .2
The probability that the bag of 36 golf balls will weigh less than 360 grams is the same as the
probability that the mean or average weight of the 36 golf balls will be less than 10 grams (360/36 =
10).
P(mean of X <10) = P((mean of X- 9.5)/.2 < (10 - 9.5)/.2) = P(Z<(.5)/.2) =
P(Z<2.5) = .9938
(a) 0.9938
(b) 0.0062
(c) 0.3372
(d) 0.6628
Correct answer is (a) 0.9938.
4