Download GTS 111 Practice Final Exam

GTS 111 Practice Final Exam
By Dr. Nirattaya Khamsemanan
Disclaimer: this practice exam is my attempt to help you study for the exam. Things in here might
not be in the real exam and vice versa. Don’t use this as your main study. There might be some typos
and/or mistakes.
1. Rolling a pair of two fair distinct dice
(a) Find the probability distribution of the sum of the the upper faces appearing when the two
dice are rolled. What is the random variable in this case?
Answer: X = sum of the upper faces.
k
2
3
4
5
6
7
8
9
10
11
12
f (k) = p(X = k)
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
(b) Find the probability distribution of the number of the upper faces divisible by 2 appearing
when the two dice are rolled. What is the random variable in this case?
Answer: X = the number of the upper faces divisible by 2
k
0
1
2
f (k) = p(X = k)
9/36
18/36
9/36
(c) Find the probability distribution of the number of fives appearing when the two dice are rolled.
What is the random variable in this case?
Answer: X = the number of fives appearing when the two dice are rolled.
k
0
1
2
f (k) = p(X = k)
25/36
10/36
1/36
2. A couple plan to have 4 children. Assuming that they will have one child at the time.
(a) Find the probability distribution of the total number of daughters that this couple could have.
1
Answer: X = the total number of daughters.
k
0
1
2
3
4
f (k) = p(X = k)
1/16
4/16
6/16
4/16
1/16
(b) Draw a histogram for the probability distribution of the total number of daughters that this
couple could have.
(c) Is this probability distribution a uniform distribution? Explain your reason.
Answer: No
(d) Is this probability distribution a binomial distribution? Explain your reason.
Answer: Yes
(e) What is the probability that there are two boys and two girls?
Answer:
p(2 successes from 4 trials) = C24
1 4
2
(f) What is the probability that there are three children of one gender and one of the other?
Answer:
p(are three children of one gender and one of the other) = C14
1 4
2
+ C34
1 4
(g) What is the probability that all four children are of the same gender?
1 4
1 4
+ C44
p(all four children are of the same gender) = C04
2
2
(h) what is the probability that there are at least one girl?
p(at least one girl) = 1 − p(no girls) = 1 − C04
1 4
2
3. Check whether the function given by
f (k) =
(k − 2)2 − 1
for k = 0, 1, 2, 3, 4
9
can serve as the probability distribution of a discrete random variable.
Answer: No since f (2) < 0.
4. The cumulative distribution (distribution






F (x) =





function) of a random variable X is given by
0
1/5
1/3
7/15
1
2
for
for
for
for
for
k<0
0≤k<3
3≤k<4
4≤k<6
k≥6
2
(a) Draw a graph of the cumulative distribution of this random variable.
(b) Find p(X ≤ 5)
Answer:
p(X ≤ 5) = F (5) =
7
15
(c) Find p(2 < X ≤ 4)
Answer:
p(2 < X ≤ 4) = F (4) − F (2) =
7
1
−
15 5
(d) Find p(X = 3)
Answer:
p(X = 3) = F (3) − F (2) =
1 1
−
3 5
(e) Find the probability distribution of X where k = 0, 1, 2, 3, ..., 7.
f (0)
f (1)
f (2)
f (3)
f (4)
f (5)
f (6)
f (7)
1
5
= F (1) − F (0) = 0
= F (0) =
= F (2) − F (1) = 0
2
= F (3) − F (2) =
15
2
= F (4) − F (3) =
15
= F (5) − F (4) = 0
8
= F (6) − F (5) =
15
= F (7) − F (6) = 0
5. The number of cars waiting to be served at a gas station was counted at the beginning of every
minutes during the morning rush hour. The frequency distribution is given in the following table.
Number of cars waiting
Number of occurrences
0
0
1
9
2
21
3
15
4
12
5
3
(a) Find the probability distribution of the number of cars waiting to be served at a gas station
in the beginning of each minute during rush hour.
Answer:
k
0
1
2
3
4
5
f (k) = p(X = k)
0
9/60
21/60
15/60
12/60
3/60
(b) Draw the histogram of this probability distribution.
3
(c) What is the mean, median and the mode of the number of cars waiting to be served at a gas
station in the beginning of each minute during rush hour.
Answer: Median is 2.5, Mode is 2
Pn
i=1
Mean =
xi fi
n
159
= 2.65
60
=
(d) What is the standard deviation of the number of cars waiting to be served at a gas station in
the beginning of each minute during rush hour.
Answer:
Pn
Variance
=
Standard deviation
=
i=1 (xi
√
− x)2 fi
n
0.81 = 0.9
=
48.6
= 0.81
60
6. Forty percent of a particular model of car are silver. What is the probability that in the next 10
observations of this model you observe at least 5 silver cars.
Answer: This is the binomial experiment with n = 10 and p(S) = 0.4.
p(at least 5 silver cars) = 1 −
4
X
b(i, 10, 0.4)
i=0
7. Which of the following is most likely: ”getting at least one three in six rolls of a die” or ”getting at
least two threes in twelve rolls of a die”?
Answer:
Getting at least one three in six rolls of a die: This is the binomial experiment with n = 6
and p(S) = 1/6.
p(at least one three in six rolls) = 1 − b(0, 6, 1/6) = 0.665102
Getting at least two threes in twelve rolls of a die: This is the binomial experiment with
n = 12 and p(S) = 1/6.
p(at least two threes in twelve rolls) = 1 − b(0, 12, 1/6) + b(1, 12, 1/6) = 0.618667
Thus, ”getting at least one three in six rolls of a die” is more likely to happen.
8. A sample of 4 men chosen at random are asked to try a new type of razor and to state whether
they prefer it to the type they are presently using. Suppose that probability is 0.1 that each man
will prefer the new type and that the men’s preferences are independent. The random variable X
represents the number of men who prefer the new razor.
(a) If possible, re-write each man’s preferences as a a Bernoulli trial. Identify the successful outcomes and p, probability of success of each trial.
Answer: Each Bernoulli trial is each man’s preference. The two outcomes are ”prefer new
razor” and ”prefer the type they use now” with ”prefer new razor” is a success on each trial.
The probability of success on each trial is p(S) = 0.1.
4
(b) Is this situation a binomial experiment? Explain your reason.
Answer: Yes because there is fixed number of trial n = 4. Each trial has two outcomes, prefer
new razor, and not. If we let the successful outcome be ”prefer new razor” with p(S) = 0.1.
Each trial is independent.
(c) Find the probability that only the first and the last men like this new type of razor.
Answer:
p(SF F S) = (0.1)(0.9)(0.9)(0.1) = 0.0081
(d) Find the probability that only two men like this new type of razor.
Answer:
p(2 successes in 4 trials) = b(2; 4, 0.1) = C24 (0.1)2 (0.9)2
(e) Find the probability that at most two men like this new type of razor.
Answer:
p(at most two men like this new type of razor) = b(0; 4, 0.1) + b(1; 4, 0.1) + b(2; 4, 0.1)
(f) Write the probability distribution defining this experiment.
Answer:
p( k successes in 4 trials) = b(k; 4, 0.1) = Ck4 (0.1)k (0.9)4−k
(g) Construct a table for this probability distribution.
Answer:
k
0
1
2
3
4
f (k) = p(X = k)
C04 (0.1)0 (0.9)4 = 0.6561
C14 (0.1)1 (0.9)3 = 0.2916
C24 (0.1)2 (0.9)2 = 0.0486
C34 (0.1)3 (0.9)1 = 0.0036
C44 (0.1)4 (0.9)0 = 0.0001
(h) Draw a histogram for this binomial distribution.
(i) Find the mean and the standard deviation for this distribution.
Answer:
µ = np = 4(0.1) = 0.4
p
p
σ =
np(1 − p) = 4(0.1)(0.9) = 0.6
(j) Find the probability that the number of successes lies within 2 standard deviation of the mean.
Answer: Since the mean of the successes is 0.4. Therefor the number of successes lies within
2 standard deviation of the means are 0 and 1. Thus the probability that the number of
successes lies within 1 standard deviation of the mean is
5
p(0 successes in 4 trials) + p(1 successes in 4 trials)
= C04 (0.1)0 (0.9)4 + C14 (0.1)1 (0.9)3
= 0.9477
(k) Find the cumulative distribution of the total number of men who like this new type of razor.
Answer:

0




0.6561



0.9477
F (k) =
 0.9963



 0.9999


1
for
for
for
for
for
for
k<0
0≤k<1
1≤k<2
2≤k<3
3≤k<4
k≥4
(l) Draw a graph of the cumulative distribution the total number of men who like this new type
of razor.
9. Use the binomial formula to expand (t − 1)6
t6 +C16 t5 (−1)+C26 t4 (−1)2 +C36 t3 (−1)3 +C46 t2 (−1)4 +C56 t(−1)5 +(−1)6 = t6 −6t5 +15t4 −20t3 +15t2 −t+1
10. Use the binomial formula to find the fourth term in the expansion of (x + 3)20
C320 x17 33
11. A salesperson determines that the probability of making a sale to a customer is 0.02. He plans to
approach 80 customers this week. Use Poisson distribution to approximate binomial distribution
to find the probability that this salesperson will make any sale at all this week.
Answer: p(making any sale) = 0.798103
12. The number of gene mutation under a given level of radiation is a random variable having a Poisson
Distribution with λ = 3.
(a) What is the mean and the standard deviation of the number of gene mutation under this given
level of radiation?
Answer:
Mean is λ = 3
Variance is λ = 3
√
√
Standard Deviation is λ = 3 = 1.732
(b) What is the probability of the occurrence of at least 1 gene mutation?
Answer: p(at least 1 gene mutation) = 1 −
30 e−3
0!
(c) What is the probability of the occurrence of exactly 5 gene mutation?
Answer: p(exactly 5 gene mutation) =
35 e−3
5!
(d) What is the probability of the occurrence of at most 2 gene mutation?
Answer: p(at most 2 gene mutation) =
30 e−3
0!
+
31 e−3
1!
+
32 e−3
2!
13. Sketch the appropriate picture and find the probabilities indicated below of the standard normal
distribution.
6
±J
•. _._._.•.__ .•..... __ ..•_..__....
.
valu e of z for whi ch the
.0802 .
F ig. 26(b) . Expr ess t he
of standard deviati ons
... __._1-_ ...
..
..
• _. .. .. .• •_
~~
J1 = 9 0
I
(a) p(z > 0.6)
-z
90
z
Answer: (a)
(c) 94
(d) 104
(b)
Figure 26
p(z > 0.6)
=
1 − A(0.6)
=
1 − 0.7257
(b) p(z < −1.05)
r A( z) (Ta ble 2) to find
der the standard normal
Ll:k
il
9. Area is .0401.
Answer:
~
p(z < −1.05)
5
= A(−1.05)
=
0.1469
z
(c) p(−0.8 < z < −0.35)
10. Ar ea is .0456.
Answer:
£~
p(−0.8 < z < −0.35)
~
il
<1 = 8
I
1
= A(−0.35) − A(−0.8)
=
0.3632 − 0.2119
z area of the shaded region under the standard normal curve is as
-zof z for which the
14. Find the value
specified.
1 1.Area
Ar eaisis0.5468
.5468 .
(a)
~
5
of z for which the area
n dard no rm al curv e is as
-z
z
388
Chapter 7 Probability and Statistics
Answer: z = 0.75
(b)12.
Area
is is
0.6915
Ar ea
.6915.
24 .
11
= 30
~
32
-z
25.
~
13.
What is
the
Answer:
z=
0.590th percen t ile of t he sta nd ard norma l
d istributi on ?
G distributed
10 with
15. Assume
that
scores for of
a first-year
classnorma
at a l university are normally
14. W
hat SAT
is t heverbal
65t h percentile
t he standard
mean 520
andutstandard
deviation 75.
ion ?
distrib
26.
(a)InThe
top 10%
of the
students
into the honors program for English. What is the
Exercises
15- 18,
determ
in e f1 anare
d (J placed
by inspection.
lowest score for admittance into the honors program?
15.
(b) Find the 98th percentile of SAT verbal scores.
5 6
Answer: X =4 673.75
16.
7
8
~
7
60
70
80
~
- -- -
Answer: X = 616.09
90
100
3.5
4.5
27. What is t he pro babilit y t hat an outcome of a narc:
ra nd om var ia ble is within t wo standard deviati ons
t he mean ?
28. W hat is t he prob ability t ha t an outcome of a nOD
rand om va riab le is within t hree standard devi ations
t he mean ?
17.
29. Find t he value of (J for a nor mal random variabl­
having f1 = 5, if P r (X S 6) = .9772.
7
16. Let X be the amount of soda released by a soft-drink dispensing machine into a 6-ounce cup. Assume that X is normally distributed with σ = 0.25 ounces and that the average fill can be set
by the vendor. At what quantity should the average fill be set so that no more than 0.5% of the
releases overflow the cup?
Answer: µ = 5.35
17. Laboratory mice are given an illness for which the usual recovery rate is 1/6. A new drug is tested
on 20 of the mice, and 8 of them recover. Use Normal approximation to the binomial distribution to
find the probability that 8 or more would have recovered if the 20 mice had not been given the drug.
Answer: p(X > 7.5) = 0.0062 So this drug actually have some effects.
18. The incidence of color blindness among the men in a certain country is 33%. 50 men were randomly
chosen. Use the Normal approximation to binomial distribution to find
µ = 16.5, σ = 3.325
(a) the probability of finding more than 16 color-blind men.
Answer: p(X > 16.5) = 0.5
(b) the probability of finding less than 22 color-blind men.
Answer: p(X < 21.5) = 0.9332
(c) the probability of finding at least 12 color-blind men.
Answer: p(X > 11.5) = 1 − 0.0668
(d) the probability of finding at most 26 color-blind men.
Answer: p(X < 26.5) = 0.9987
(e) the probability of finding exactly 15 color-blind men.
Answer: p(14.5 < X < 15.5) = 0.3821 − 0.2743
8