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HONR 399
October 29, 2010
Chapter 27 Answers
In every answer, we use Z ∼ N (0, 1) as a standard normal random variable.
1. The quantity of sugar X (measured in grams) in a randomly-selected piece of candy is
normally distributed, with expected value E[X] = µ = 22 and variance Var(X) = σ 2 = 8.
Find the probability that a randomly-selected piece of candy has less than 20 grams of sugar.
We have
X − 22
20 − 22
√
P (X < 20) = P
< √
8
8
≈ P (Z < −.71)
= P (Z > .71)
= 1 − P (Z ≤ .71)
≈ 1 − .7611
= .2389
2. Consider a normal random variable X with expected value µ and standard deviation σ.
Find the probability that X is within one standard deviation of its expected value:
X −µ
µ+σ−µ
µ−σ−µ
≤
≤
P (µ − σ ≤ X ≤ µ + σ) = P
σ
σ
σ
= P (−1 ≤ Z ≤ 1)
= P (Z ≤ 1) − P (Z ≤ −1)
= P (Z ≤ 1) − P (Z ≥ 1)
= P (Z ≤ 1) − (1 − P (Z ≤ 1))
≈ .8413 − (1 − .8413)
= .6826
Find the probability that X is within two standard deviations of its expected value:
µ − 2σ − µ
X −µ
µ + 2σ − µ
≤
≤
P (µ − 2σ ≤ X ≤ µ + 2σ) = P
σ
σ
σ
= P (−2 ≤ Z ≤ 2)
= P (Z ≤ 2) − P (Z ≤ −2)
= P (Z ≤ 2) − P (Z ≥ 2)
= P (Z ≤ 2) − (1 − P (Z ≤ 2))
≈ .9772 − (1 − .9772)
= .9544
1
Find the probability that X is within three standard deviations of its expected value:
X −µ
µ + 3σ − µ
µ − 3σ − µ
≤
≤
P (µ − 3σ ≤ X ≤ µ + 3σ) = P
σ
σ
σ
= P (−3 ≤ Z ≤ 3)
= P (Z ≤ 3) − P (Z ≤ −3)
= P (Z ≤ 3) − P (Z ≥ 3)
= P (Z ≤ 3) − (1 − P (Z ≤ 3))
≈ .9987 − (1 − .9987)
= .9974
3. Assume that the annual precipitation in West Lafayette, Indiana, is normally distributed,
with expected value µ = 36.3 inches and variance σ 2 = 8.41. A rare species of frog lives in
West Lafayette. This rare species of frog is known to reproduce during the year only if the
annual precipitation is between 35 and 39 inches. What is the probability that the species
of frog is able to reproduce this year?
Let X denote the annual precipitation this year. Then
35 − 36.3
X − 36.3
39 − 36.3
√
P (35 ≤ X ≤ 39) = P
≤ √
≤ √
8.41
8.41
8.41
≈ P (−.45 ≤ X ≤ .93)
= P (X ≤ .93) − P (X ≤ −.45)
= P (X ≤ .93) − P (X ≥ .45)
= P (X ≤ .93) − (1 − P (X ≤ .45))
≈ .8238 − (1 − .6736)
= .4974
4. Assume that the height of an American female is normal with expected value µ = 64 and
standard deviation σ = 12.8.
What is the probability that an American female’s height is 66 inches or taller?
Let X denote the height of an American female. Then
X − 64
66 − 64
P (X ≥ 66) = P
≥
12.8
12.8
≈ P (Z ≥ .16)
= 1 − P (Z ≤ .16)
≈ 1 − .5636
= .4364
2
The heights of 10 American females are measured (in inches). Let Y be the number of
the 10 females whose height is 66 inches or taller. Find P (Y = 7).
We note that Y is Binomial with n = 10 and p = .4364. So
n 7
10
3
P (Y = 7) =
p (1 − p) =
(.4364)7 (1 − .4364)3 = 0.06476.
7
7
5. Review question: If X is a continuous uniform random variable on the interval [a, b],
find the kth moment of X, i.e., find E[X k ].
We can directly compute
k
Z
b
E[X ] =
a
b
bk+1 − ak+1
1
xk+1 =
x
dx =
b−a
k + 1 x=a (k + 1)(b − a)
k
Find the variance of X, i.e., Var(X).
The expected value of X is
b1+1 − a1+1
b 2 − a2
(b + a)(b − a)
E[X] = E[X ] =
=
=
= (a + b)/2.
(1 + 1)(b − a)
2(b − a)
2(b − a)
1
The second moment of X is
E[X 2 ] =
b 3 − a3
(b − a)(a2 + ab + b2 )
b2+1 − a2+1
=
=
= (a2 + ab + b2 )/3.
(2 + 1)(b − a)
3(b − a)
3(b − a)
The expected value of X is
Var(X) = E[X 2 ] − (E[X])2
2
a2 + ab + b2
a+b
−
=
3
2
2
2
4a + 4ab + 4b − 3a2 − 6ab − 3b2
=
12
a2 − 2ab + b2
=
12
(b − a)2
=
12
3