作業九 (week11 習題：5.6, 5.8, 5.9, 5.15, 5.16, 5.18, 5.19, 5.23, 5.32, 5.38)
5.6
a.
b.
0, x  0
 x
F ( x)   0.25dx  0.25 x,0  x  4
0
1,4  x

c. P(X < 1) = 0.25
d. P(X < 0.5) + P(X > 3.5) = P(X < 0.5) + 1 – P(X < 3.5) = 0.25
5.8
a. P(380 < X < 460) = P(X < 460) – P(X < 380) = 0.6 - 0.4 = 0.2
b. P(X < 380) < P(X< 400) < P(X < 460)→0.4 < P(X < 400) < 0.6
5.9
Let X = Total cost for production process
Y = The number of units produced
X=1000+2Y,E(Y)=500,Var(Y)
E(X)=E(1000+2Y)=1000+2E(Y)=2000
Var(x)=Var(1000+2Y)=4Var(Y)=3600
5.15
Let Y= The money employees are paid
X= The amount of money generated in a week
Y=60+0.2X,E(X)=700,Var(X)= 130 2
E(Y)=60+0.2E(X)=200
Var(Y)=0.04Var(X)= (0.2  130) 2
 Y  0.2 130  26
5.16
Let Y= The annual salary a salesperson receive
X= The annual value of the order
Y=6000+0.08X,E(X)=600000,  X  180000
E(Y)= 6000 + 0.08E(X)=54,000
 Y =|0.08|  X = 0.08(180000) = 14400
5.18
a. Find Z0 such that P(Z < Z0) = .7, closest value of Z0 = .52
b. Find Z0 such that P(Z < Z0) = .25, closest value of Z0 = -.67
c. Find Z0 such that P(Z > Z0) = .2, closest value of Z0 = .84
d. Find Z0 such that P(Z > Z0) = .6, closest value of Z0 = -.25
5.19
X follows a normal distribution with µ = 50 and 2 = 64
a. Find P(X > 60). P(Z >
60  50
) = P(Z > 1.25) = .5- .3944 = .1056
8
b. Find P(35 < X < 62). P(
35  50
62  50
<Z<
) = P(-1.88 < Z < 1.5)
8
8
= .4699 + .4332 = .9031
c. Find P(X < 55). P(Z <
55  50
) = P(Z < .62) = .5+ .2324 = .7324
8
d. Probability is .2 that X is greater than what number?
Z = .84.
.84 
X  50
;
8
X = 56.72
e. Probability is .05 that X is in the symmetric interval about the mean between
which two numbers?
.06 
Z = +/- .06.
X  50
.
8
X = 49.52 and 50.48.
5.23
a. P(Z >
b. P(
1, 000  1, 200
) = P(Z > -2) =FZ(2) = .9772
100
1,100 1, 200
1,300  1, 200
<Z<
) = P(-1 < Z < 1) = 2FZ(1) –1 = .6826
100
100
c. P(Z > 1.28) = .1,
plug into the Z-formula all of the known information and solve for the unknown:
1.28 =
X  1, 200
.
100
Solve algebraically for X. Therefore, X = 1,328
5.32
For Investment A, the probability of a return higher than 10%:
P(Z >
10  10.4
) = P(Z > -.33) = FZ(.33) = .6293
1.2
For Investment B, the probability of a return higher than 10%
P(Z >
10  11.0
) = P(Z > -.25) = FZ(.25) = .5987
4
Therefore, Investment A is a better choice
5.38
P(14 < X < 16) = P (
14−15
2
<Z<
16−15
2
) = P(−0.5 < Z < 0.5) = 0.38