STAT 571 Midterm: STAT 571 UW Madison Fall 03Mid1 Solutions
Stat/For/Hort 571
Midterm I, Fall 2003
Brief Solutions
1. (a) A stem-and-leaf display:
0 | 4
0 | 78
1 | 1344
1 | 67
2 | 1
(b) Since y0.25 = 8 and y0.75 = 16, IQR = 8.
2. Let Adenote the event that Habibi barks and
let Bdenote the event that Spotty barks. We
have P(A) = 0.5 and P(B) = 0.7.
(a) The event that both dogs bark corresponds
to the event of “Aand B”. By indepen-
dence, P(Aand B) = P(A)×P(B) = 0.35.
(b) The event that at least one dog barks cor-
responds to the event of “Aor B”. Thus
P(Aor B) = P(A)+P(B)−P(Aand B) =
0.5 + 0.7−0.35 = 0.85.
(c) The probability distribution of Yis:
y0 1 2
p(y) 0.15 0.5 0.35
Note that from (a), P(Y= 2) = 0.35,
P(Y= 1) = P(Y≥1) −P(Y= 2) =
0.85 −0.35 = 0.5, and P(Y= 0) =
1−P(Y≥1) = 1 −0.85 = 0.15. Thus
E(Y) = 0 ×0.15 + 1 ×0.5 + 2 ×0.35 = 1.2
and V ar(Y) = (0 −1.2)2×0.15 + (1 −
1.2)2×0.5 + (2 −1.2)2×0.35 = 0.46.
3. (a) Since V2=(10 −1)S2
9∼χ2
9,
P(8.34 ≤S2≤14.68)
=P(10 −1)8.34
9≤V2≤(10 −1)14.68
9
=P(8.34 ≤V2≤14.68)
=P(V2≥8.34) −P(V2≥14.68)
= 0.5−0.1 = 0.4.
(b) Since ¯
Y∼N(20,9
10 ), Z=¯
Y−20
r9
10
∼
N(0,1). We know that P(Z≤1.645) =
0.95, thus
0.95 = P
¯
Y−20
q9
10
≤1.645
=P(¯
Y≤1.645 ×r9
10 + 20)
and y∗= 1.645 ×q9
10 + 20 = 21.56.
(c) False. Since V ar(¯
Y) = 9
n, by the same
argument as in (b), y= 1.645 ×q9
n+ 20.
The value yis smaller than y∗, because
the variance V ar(¯
Y) is smaller, when the
sample size nis larger than 10.
4. (a) Let Ydenote the number of infected plants
among the 15 plants. Then Yis binomial
with n= 15 and p= 0.1. The probability
that at least 1 plant is infected is
P(Y≥1) = 1 −P(Y < 1)
= 1 −P(Y= 0)
= 1 −(0.9)15 = 0.794.
Note that by using the complement of the
event of interest, we could avoid comput-
ing P(Y≥1) = P(1) + P(2) + ···+P(15).
(b) Here Wis binomial with n= 150 and p=
0.1. The mean and variance of Ware np =
15 and np(1 −p) = 13.5. Since np = 15 ≥
5 and n(1 −p) = 135 ≥5, we can use
WNA ∼N(15,13.5). Since
0.90 = P(W≥a)
≈P(WNA ≥a)
=PWNA −15
√13.5≥a−15
√13.5
=P(Z≥a−15
√13.5).
and P(Z≥ −1.282) = 0.90, we set
a−15
√13.5=−1.282. Thus a= 15 −1.282 ×
√13.5 = 10.29, or a= 10.
Grade Distribution
100:21
90-99:59
80-89:22
70-79:19 mean = 88, median = 92
60-69:7
50-59:4
Document Summary
2 | 1 (b) since y0. 25 = 8 and y0. 75 = 16, iqr = 8: let a denote the event that habibi barks and let b denote the event that spotty barks. We have p (a) = 0. 5 and p (b) = 0. 7. (a) the event that both dogs bark corresponds to the event of a and b . By indepen- dence, p (a and b) = p (a) p (b) = 0. 35. (b) the event that at least one dog barks cor- responds to the event of a or b . P (a or b) = p (a)+p (b) p (a and b) = 0. 5 + 0. 7 0. 35 = 0. 85. (c) the probability distribution of y is: y p(y) Since v ar( y ) = 9 n , by the same argument as in (b), y = 1. 645 q 9 n + 20.
