Solutions: Homework #2
1a)
P(A) = 5/9 = .56
P(B) = .2/9 = .22
P(A  B) = P(A) + P(B) - P(A  B) = .56 + .22 -.11 = .67
P(A  B) = P(A)*P(B|A) = .56*(.2) = .11
P(A|B) = .5
P(B|A) = .2
b)
A and B are not independent because P(A|B) is different from P(A). In other
words, the probability that a person is female changes if we know that their name
begins with the letter J.
2) Using Bayes Theorem (+ = Positive test; D = Drunk):
P(D|+) =
P(+|D)*P(D)
P(+|D)*P(D) + P(+|D`)*P(D`)
=
=
(.9)*(.2) / [(.9)*(.2) + (.05)*(.80)]
.18 / (.18 + .04)
=
.18 / .22
3)
Siblings
Frequen
cy
Percent
Valid
Cumulative
Percent
Percent
Valid 0
9
6.9
6.9
6.9
1
65
50.0
50.0
56.9
2
36
27.7
27.7
84.6
3
11
8.5
8.5
93.1
4
7
5.4
5.4
98.5
8
1
.8
.8
99.2
11
1
.8
.8
100.0
130
100.0
100.0
Total
Statistics
Siblings
N
Valid
Missing
Mean
130
0
1.67
Std. Deviation
1.372
Variance
1.882
.81
Above, you can find the SPSS output. I’m going to transform it a little to show how to
calculate E(x) and S using the discrete random variable formula.
P(x) rounded
x*p(x)
[x-E(x)]2 [(x-E(x))^2]*p(x)
x
frequency
(f/116)
0
9
.069
0.00
2.81
0.19
1
65
.500
0.50
0.46
0.23
2
36
.277
0.55
0.1
0.03
3
11
.085
0.26
1.75
0.15
4
7
.054
0.22
5.40
0.29
8
1
.008
0.06
39.98
0.32
11
1
.008
0.09
86.92
0.70
1.91
1.68
E(x)
Var
SD
Sum [x*p(x)]
Sum[x-E(x)^2*p(x)]
sqrt(var)
1.68
1.91
1.38
You will notice that the values for SPSS differ from those I calculated below. The reason
for that is the SPSS values are based on the actual number of observations that we have.
The discrete random variable calculation assumes we are dealing with an infinite number
of families/siblings. Please come see me if you would like more details about this.
4)
n = # of trials = 3
x = # of successes = 2
p = P(success) = .4
1-p = P(failure) = .6
Binomial formula: (n choose x) * px * (1-p)n-x
P(x=2) = (3 choose 2) * .42 * .61
=
3 * .16 * .6
5)
N
r
n
x
=
.288
# of marbles = 10
# of successes (blue marbles) = 4
total # of marbles picked = 3
# of blue marbles picked = 1
Hypergeometric Rule: (r choose x) * [(N-r) choose (n-x)] / (N choose n)
P(x=1)
=
(4 choose 1) * [(10-4) choose (3-1)] / (10 choose 3)
=
(4 choose 1) * (6 choose 2) / (10 choose 3)
=
4
*
15
/
120
=
60 / 120
=
.50
6)
Statistics
Height
N
Valid
Missing
130
0
Mean
67.84
Std. Deviation
4.426
Variance
z
7)
19.590
=
x - mean / s
=
63 – 67.8 / 4.4
=
-4.8 / 4.4
=
-1.09
According to the Z-table, .8621 of the standard normal curve lies above –1.09 (Body).
This suggests that the proportion of the AD student body that is over 5'3" is .8621.
x - mean / s
64 – 67.8 / 4.4
-3.8 / 4.4
-0.87




z
z
z
z




x - mean / s
72 – 67.8 / 4.4
4.2 / 4.4
0.94
According to the table, .8078 of the standard normal curve lives above –0.87 (Body),
and .1736 lies above 0.94 (Tail). Therefore, the area between these two points is
.8078 -.1736 = .6342. This suggests that the proportion of AC students that fall
between 5'4" and 6' is .6342. Alternatively, you could have calculated the area in the
tail below -0.87 (.1922) and the area in the body beneath 0.94 (.8264; .8264 -.1922 =
.6342. OR, you could have found the area between 0 and -0.87 (.3078) and added that
to the area between 0 and 0.94 (.3264). The sum of .3078 and .3264 is .6342. Golly,
there are a lot of ways home for this one.