Stat/For/Hort 571 — Midterm I, Fall 98 —
Brief Solutions
1. (a) One helpful display is the stem and leaf display. For this
problem it is good to use the first digit as the stem, but
stretch the stem out by putting numbers with “leaves”
that are <50 with one stem and those ≥50 with another
stem, as in the following:
3 | 35
3 | 64, 84, 99
4 | 23, 11, 18, 02, 30
4 | 83, 77, 91, 59, 68, 76
5 | 04, 20, 43
5 | 61
The display is roughly symmetric and the center is
around 450cm.
(b) Let µb e the mean height of this variety of seedlings.
Thus we have
H0:µ= 420 cm and HA:µ6= 420 cm
We calculate ¯y= 8548/19 .
= 449.89 and we are given
σ2= 3800.So
z=¯y−µ
σ/√n=449.89 −420
√3800/√19
.
= 2.11.
The p-value is 2P(Z≥2.11) = 2(0.0174) = 0.0348.
There is moderate evidence against H0,and we reject
the null at 10% and 5% since the p-value falls below
these levels. However, we do not reject H0at 1%.
2. (a) The question asks for the following shaded area:
Z
-3 -2 -1 0 1 2 3
0.0214
From the tables, we find that P(Z≥2) = 0.0228 and
P(Z≥3) = 0.0013.Thus P(2 ≤Z≤3) = 0.0228 −
0.0013 = 0.0215.The desired area is twice this and is
0.043.
(b) We have that ¯
W∼N(µ, σ2/n) = N(85,60/16) .
=
N(85,1.9362).
79.192 83.064 86.936 90.808
0.2501
Z
-3 -2 -1 0 1 2 3
0.2501
The desired w∗is 0.67 standard deviations away from
the mean since P(Z≥0.67) = 0.25,and we solve
w∗−85
1.936 = 0.67
so that w∗= 85 + (0.67)(1.936) .
= 86.3.
3. (a) Since Y∼B(8,0.3),we have that
P(Y= 0) + P(Y= 1)
=8!
0!8! 0.300.78+8!
1!7! 0.310.77
.
= 0.0576 + 0.198 .
= 0.256.
(b) Here Y∼B(80,0.3).We appeal to the normal approxi-
mation, noting that np = 24 and n(1 −p) = 56 so that
this is appropriate. Thus YNA ∼N(np, np(1 −p)) .
=
N(24,4.0992),and the desired probability is P(Y≤
10) .
=P(YNA ≤10) = P(Z≤(10−24)/4.099) = P(Z≤
−3.41) = 0.0003.
(c) When nis large, the proportion Y /n becomes closer to p
(with a standard deviation equal to pp(1 −p)/n. Thus,
it is much less likely to get a proportion that is less than
or equal to 0.125 (= 1/8 = 10/80).
4. (a) E(Y) = µ=Pyp(y) = 0 ×0.2+1×0.5 + 5 ×0.3 = 2
and Var(Y) = σ2=P(y−µ)2p(y) = (0 −2)2×0.2 +
(1 −2)2×0.5 + (5 −2)2×0.3 = 4.
(b) By the Central Limit Theorem (CLT) (see part (c)), we
have that ¯
Yis approximately N(µ, σ2/n) = N(2,0.22).
Thus P(¯
Y < 1.5) .
=P(Z < (1.5−2)/0.2) = P(Z <
−2.5) = P(Z > 2.5) = 0.0062.
(c) To calculate the probability in (b) we need to appeal to
the CLT which says that ¯
Yfor a random sample is ap-
proximately normal provided that the sample size nis
large. How large nmust be dep ends on the population
from which the observations are sampled. Typically the
approximation is reasonable if n≥30 as long as the un-
derlying population is not too skewed. The distribution
here is not too heavily skewed.
5. (a) Since n= 4, E(V2) = n−1 = 3 and Var(V2) = 2(n−
1) = 6 so that the standard deviation = √6.
= 2.45.
(b) As the mean is 3 and the standard deviation is 2.45,
we are interested in the chance that V2falls within
3±2.45 = [0.55,5.45],or P(0.55 ≤V2≤5.45),as illus-
trated in the picture below:
Using Table B, we can bound separately the two parts
that are not shaded, noting that df = n−1 = 3.
For the right side 0.1 = P(V2≥6.25) < P (V2≥
5.45) < P (V2≥4.11) = 0.25.Then for the left,
0.05 = 1 −0.95 = P(V2≤0.35) < P (V2≤0.55) <
P(V2≤0.58) = 1 −0.9 = 0.1.
The two nonshaded areas are then bounded between
0.1 + 0.05 = 0.15 and 0.25 + 0.1 = 0.35.Thus, the
shaded area is b ounded between 1 −0.35 = 0.65 and
1−0.15 = 0.85.Note that because Table B is sparse,
interpolation is dangerous. [Note an exact calculation
with Minitab gives 0.766.]
Grade Distribution
100:5
95-99:34
90-94:27
80-89:26 n = 151
70-79:29 mean = 82.98, median = 87
60-69:21 quartiles = 73, 95
50-59:6 s = 13.9
<50:3