```Unit 4 Quiz 2 Review Answers
1. a. P(X = 3) = 0.28
b. P(X ≤ 3) = 0.16 + 0.22 + 0.28 = 0.66
c. P(1&lt;X≤4) = 0.22 + 0.28 + 0.20 = 0.70
d. E(X) = 2.94 units
SD(X) = 1.2714 units
2. Probability model for the payout of a prize
X = payout
P(X=x)
-\$2
1496/1500
\$73
1/1500
\$148
1/1500
\$248
1/1500
\$498
1/1500
For (b) and (c), just put X-values in L1, probabilities in L2 and then do 1-var stats L1, L2:
b. E(X) = -\$1.35
c. SD(X) = \$15.06
d. E(X + X + X) = -1.35 + -1.35 + -1.35 = -\$4.05
SD(X + X + X) =
15.062  15.062  15.062  \$26.08
3. SAT Scores
M – score for a randomly selected male
F – score for a randomly selected female
E(M) = 1532
E(F) = 1506
SD(M) = 312
SD(F) = 304
a. M + F;
E(M + F) = 1532 + 1506 = 3038 points
SD(M + F) =
3122  3042  435.61 points
b. M – F
E(M – F) = 1532 – 1506 = 26 points
SD(M – F) =
3122  3042  435.61 points
c. Assume a normal model with M – F ~ N(26, 435.61).
Female scored higher than male  P(F &gt; M)  P(M -- F &lt; 0)
P(M –F &lt; 0) = normalcdf(-E99, 0, 26, 435.61) = 0.4762
There is a 47.62% chance that a randomly selected female scored higher than a randomly selected male on the
SAT.
4. Random Variables
a. -2X
E(-2X) = -2(12) = -24
SD(-2X) = 2(5) = 10
b. 4Y – 7
E(4Y – 7) = 4(18) – 7 = 65
SD(4Y – 7) = 4(8) = 32
c. X + Y
E(X + Y) = 12 + 18 = 30
d. X – Y
E(X – Y) = 12 – 18 = -6
SD(X + Y) =
52  82  9.43
e. X1 + X2
E(X1 + X2) = 12 + 12 = 24
SD(X1 + X2) =
SD(X – Y) =
X
Y
Mean
12
18
SD
5
8
52  82  9.43
f. 2X – 4Y
E(2X – 4Y) = 2(12) – 4(18) = -48
52  52  7.07
E(2X – 4Y) =
22 (52 )  42 (82 )  33.53
5. The probability distribution below represents (X) the length of long distance calls in minutes.
X
P(X)
5
.1
10
.2
15
.3
20
.3
25
.1
a. What is the expected length of a long distance call? What is the standard deviation?
E(X) = 15.5 mins
SD(X) = 5.679 mins
b. Suppose that there is an initial connection charge of \$.60 and an additional charge of \$.015 for each minute.
What is the expected cost and standard deviation of the cost of a long distance call?
E(0.60 + 0.015X) = 0.60 + 0.15(15.5) = \$0.8325
SD(0.60 + 0.015X) = SD(0.015X) = \$0.0852
c. The expected cost of sending a text message is \$.20 with a standard deviation of \$.10. What is the mean
and standard deviation for the cost of making a long distance call and sending a text message?
E(C) = \$0.8325
E(T) = \$0.20
E(C + T) = \$1.0325
SD(C) = \$0.0852
SD(T) = \$0.10
SD(E + T) = \$0.1314
d. Which costs more, the call or the text? How much more? With what standard deviation?
E(C – T) = \$0.6325
Calls cost more, on average \$0.6325 with std. dev of \$0.1314
SD(E – T) = \$0.1314
6. Check: Bernoulli?
1. 2 Outcomes – has jumper cables or doesn’t have jumper cables
2. p = 0.40 and p is constant
3. 10% Condition – we can assume independence as long as the number of drivers asked is less than 10% of all
drivers.
a. G(0.40)
b. G(0.40)
c. Geom(0.40)
E(X) =
P(X = 7) = geometpdf(0.40, 7) = 0.0187
P(X &lt; 5) = geometcdf(0.40, 5) = 0.92224
1
1

 2.5 drivers
p 0.40
d. B(8,0.40)
P(X = 3) = binompdf(8, 0.4, 3) = 0.2787
e. B(6,0.40)
P  X  4  1 P( X  3 )  1 binomcdf ( 6,0.4,3 )  0.1792
f.
B(10,0.40)
P  X  6  1 P( X  5 )  1 binomcdf (10,0.4,5 )  0.1662
g. B(12,0.40)
E(X) = np = 12(0.40) = 4.8 drivers
h. B(80, 0.40)
i.
Check?
np &amp; nq &gt; 10
(80)(0.40) &amp; (80)(0.60) &gt; 10
ii.
N(32, 4.382)
iii.
P(X &lt; 30) = normalcdf(-E99, 30, 32, 4.382) = 0.324
iv.
P(42 &lt; X &lt; 61) = normalcdf(42, 61, 32, 4.382) = 0.01124
YES!
7. p = 0.15
a. B(6, 0.15)
P(X = 2) = 0.1762
b. B(8, 0.15)
P(X &gt; 3) = 1 – P(X &lt; 2) = 0.1052
c. G(0.15)
P(X = 4) = 0.09212
d. G(0.15)
P(X = 5 U 6) = P(X = 5) + P(X = 6) = 0.0783 + 0.0666 = 0.1449
e. B(10, 0.15
P(X &lt; 2) = P(X &lt; 1) = 0.5443
f. B(12, 0.15)
E(X) = 1.8 passes
SD(X) = 1.237 passes
g. G(0.15)
E(X) = 6.667 passes
h. B(25, 0.15)
P(9 &lt; X &lt; 14) = P(10 &lt; X &lt; 13) = P(X &lt; 13) – P(X &lt; 9) = 0.00214
i. B(70, 0.15)
i. Check?
np &amp; nq &gt; 10
(70)(0.15) &amp; (70)(0.150) &gt; 10 YES!
ii. N(10.5, 2.987)
iii. P(X = 20) = binompdf(70, 0.15, 20) = 0.0016
 Note: with the normal model you cannot calculate exact probabilities (ones with “equal
to”). It will give you 0 as an answer. You must go back to binompdf to calculate exact
probabilities of binomials.
iv. P(11 &lt; X &lt; 29) = normalcdf(11, 29, 10.5, 2.987) = 0.4335
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