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Practice Test for Honors Math 201
Introduction to Probability and Statistics (Liberty
University)
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E E
x
Chapter 4:
Binomial:
P( x )
=n
C( px )( qn−x)μ=np,
σ2=npq
μx
e−μ
Geometric: P(x) = qx-1p,
= 1/p
Poisson:
P( x )=
x !
¯x
±
z
c
[
σ
]
¯x
±
t
c
[
s
]
p
^
±
z
c
[
^
p
q
^
]
C.I. Formulas:
√n√n
√
n
( σ ) z
2z
2
Sample Size
Formulas:
n
≥
[
c
]
n
≥
p
^
q
^
[
c
]
1. A basketball player gets fouled attempting a three point shot. Let X be the number of
points scored on the ensuing 3 free throws. The probability distribution of X is given in
the table below: (4 pts each)
Number of points scored, X Probability of outcome, P(X)
0 0.05
1 0.20
2 0.50
3 0.25
Total 1
a) What’s the probability that X is no more than 1? 0.25
b) What’s the mean of X? 0(0.05) + 1(0.2) + 2(0.5) + 3(0.25) = 1.95
c) What’s the variance of X? points in L1; probability in L2; ox2 = variance
2. Twelve (12) people in Lynchburg are selected to serve as jurors on an especially
important case. The probability that any given juror is a senior citizen is 0.2. Let X = #
of people on the jury who are senior citizens. (5 pts each)
a) What is E(X)? 2.4
b) What is the probability that exactly 3 of the jurors will be senior citizens?
0.236 (binompdf)
c) What is the probability that less than 3 of the jurors will be senior citizens?
0.558 (binomcdf)
3. The admissions desk gets 3 phone calls per hour requested information about
Liberty. The recruiting staff takes most of these calls, except for the ½ hour where they
go to lunch, leaving only the less knowledgeable intern to handle these calls. Let X =
the number of phone calls in this ½ hour period (not a full hour) (5 pts each) (poisson)
a) What is E(X) (i.e. the mean number of phone calls in the half-hour period)? 1.5
b) What is the probability that exactly two people call during the half hour? 0.251
(use mean and number)
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c) What is the probability that two or more people call during the half hour? 0.442
4. A person is looking to replace a broken plate by bidding on matching plates on
Ebay. Let their probability of winning in any given auction be 0.15. (geomet)
a) What’s the probability that she will be able to replace the plate by winning the
fourth auction in which she bids? 0.09 (geometpdf)
b) What’s the expected number of auctions she will need to participate in
before achieving success? 1/p = 6.67
5. Monthly food expenditures (including groceries and dining out) for families of
4 are approximately normally distributed with a mean of $500 and a standard
deviation of $70. (5 pts each)
a) What’s the Z-value of 450? -0.714
b) What proportion of families of 4 spends more than $600 per month? 0.076
c) What proportion of families of 4 spends less than $450 per month? 0.238
d) What proportion of families of 4 spends between $550 and $650 on food
each month? 0.22
e) If our goal is for our food expenses to be in the bottom 10% of the
population, what is the most we can spend & still be in the bottom 10%?
$410.29
f) What is the probability that 45 randomly selected families of 4 would have
average monthly food expenses of $480 or
more? (480-500)/(70/45^1/2)
6. Matching: match the following symbols with their meanings:
h σ a) sample variance
c n b) Chi-squared random variable
i
^pc) sample size
H0d) null hypothesis
a s2e) population proportion
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f) alternate hypothesis
g) sample mean
h) population standard deviation
i) sample proportion
7. a) In a survey of 60 couples, the average number of meals eaten
outside the home per week was 2.75 meals per week. Assume that
the population standard deviation is known to be 0.75 meals per
week. Construct a 94% confidence interval for the population mean.
(5 pts) z=1.881
b) How many couples would we need to survey in order for our CI to
have a margin of error 0.10 meals per week?
199
8. The town of Salt Lick has been keeping weather records for the past
14 years. In the past 14 years, the average annual rainfall has been 30.6
inches, with a sample standard deviation of s = 2.19. Use this sample
data to construct a point estimate and a 90% margin of error for the
true average annual rainfall in Salt Lick.
STAT-TInterval; or invT to find tc
Point estimate = 30.6
Margin of error =
tc(s/n^1/5)
{29.564, 31.636}
9. a) A car dealership is trying to estimate their success rate. Last
weekend, 50 customers came to the dealership with 20 purchasing
vehicles. Use this sample to construct a 98% confidence interval for
the overall proportion of customers who purchase vehicles. (5 pts)
(1-propzint)
20/50=p=.4
0.4+2.33
12 0 12 1 12
2
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b) How many folks would they need to sample to get a 98% CI with a
margin of error (E) of 6%?
10. We have computed the following 95% Confidence Interval on a
population mean: [ 13.7, 18.2]. Given this CI, would we reject or fail
to reject the following null hypotheses?
a) H
0
: = 12 b) H
0
: = 15 c) H
0
: = 18
Test Answers
1) a) 0.05 + 0.20 = 0.25
b) 0(0.05) + 1(0.20) + 2(0.5) + 3(0.25) = 1.95
c) Var = E[X
2
] -
2
= 0
2
(0.05) + 1
2
(0.20) + 2
2
(0.50) + 3
2
(0.25) - 1.95
2
= 0.6475
2 a) 12(0.2) = 2.4
12 C3
b)
c)
(
0
.2
)
3
(
0
.
8
)
9
=
12
×
11
×
10
6
(
0
.
008
)
(
0.
13422
)
=
0
.
236
=
P
(
X
=
0
)
+
P
(
X
=
1
)
+
P
(
X
=
2
)
=
C
(
0.
2
)
0
(
0.
8
)
12
+
C
(
0
.
2
)
1
(
0
.
8
)
11
+
C
(
0
.
2
)
2
(
0
.
8
)
10
¿ 0 .0687 +0 .2062+0 . 2835=0. 5584
3 a) 3(1/2) = 1.5
b) P(2) = (1.5)2e-1.5/2! = 0.251
c) 1 - [ P(0) + P(1) ] = 1- [(1.5)1e-1.5/1! + (1.5)0e-1.5/0! ] = 1-[ 0.3347 + 0.2231] =
0.442 4) a) P(4) = (0.85)3(0.15) = 0.092 b) 1/0.15 = 6.67 or 7
5) a) z = (450−500)/70 = −0.71
b) z = 1.43; P(z > 1.43) = 1.0 − 0.9236 = 0.0764
c) z = −0.71; P(z < −0.71) = 0.2389
d) z1 = 0.71 and z2 = 2.14; P(0.71 < z < 2.14) = 0.9838 − 0.7611 = 0.2227
e) Look for a table value of 0.1000: z value is −1.28 and X = 500 −1.28(70) = 410.4
σ
X¯
f)
=70
=10 .
435
√
45
, so Z = (480 – 500)/10.435 = –1.92 and P(Z > –1.92 ) = 1 –
0.0274 = 0.9726.
50
[
(
)
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6.) h, c, i, d, a
2.
75
±
1.
96
(
0
.
75
)
=
[
2
.560
,
2
.
940
]
7a)
√60
1 . 96 ( 0 . 75 )
2
n≥
0. 10
]
=216 . 09→n=217
8. point estimate is 30.6 inches, 90% margin of error
is
0
.
40
±
2
.
326
(
√
0
.
4
(
0
.
6
)
)
=
[
0
.239
,
0
.
561
]
1. 771
2. 19
=1.
037
√14
inches
9a)
margin of error E
=
50
2.
326
(
√
0
.
4
(
0
.
6
)
)
=
0.
161
2 .326
2
n≥0. 4( 0 . 6
)
b)
0 .06
]
=360 .7 →n=361
10 a) Reject H0, b) Fail to reject H0, c) Fail to reject H0.
b)
[
