Download Math 151. Rumbos Spring 2014 1 Solutions to Assignment #22 1

Math 151. Rumbos
Spring 2014
1
Solutions to Assignment #22
1. An experiment consists of rolling a die 81 times and computing the average
of the numbers on the top face of the die. Estimate the probability that the
sample mean will be less than 3.
Solution: Let Xk , for k = 1, 2, 3, . . ., denote the outcome of the k th
roll of the die. We want to estimate Pr(X n 6 3), where X n is the
sample mean with n = 81. In this case we have µ = E(Xk ) = 3.5
for all k and σ 2 = Var(Xk ) = 35/12 for all k. Since the outcomes of
the rolls are independent and n = 81, we can use the Central Limit
Theorem to approximate
3 − 3.5
Xn − µ
√
√ 6 √ √
Pr(X n 6 3) = Pr
σ/ n
( 35/ 12)/ 81
≈ Pr(Z 6 −2.63),
where Z ∼ Normal(0, 1), by the Central Limit Theorem. Thus, using
the symmetry of the pdf of Z ∼ Normal(0, 1),
˙ 1 − 0.9957,
Pr(X n 6 3) ≈ 1 − FZ (2.63) =
so that Pr(X n 6 3) ≈ 0.0043, or about 0.43%.
2. A random sample of size 49 is taken form a distribution with mean µ and
variance σ 2 . Estimate the probability that sample mean will be within 0.7
standard deviations from the mean of the distribution.
Solution: We want to estimate Pr(|X n − µ| 6 0.7σ), where X n is
the sample mean with n = 49. Since X n is the sample mean of a
random sample and n is larger than 30, we can apply the Central
Limit Theorem to approximate
|X n − µ|
0.7σ
√
Pr(|X n − µ| 6 0.7σ) = Pr
6
σ/7
σ/ n
≈ Pr(|Z| 6 4.7),
where Z ∼ Normal(0, 1), by the Central Limit Theorem. We then
have that Pr(|X n − µ| 6 0.7σ) ≈ 1.
Math 151. Rumbos
Spring 2014
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3. A large freight elevator can transport a maximum of 9800 pounds. Suppose a
load of cargo containing 49 boxes must be transported via the elevator. Experience has shown that the weight of boxes of this type of cargo follows a
distribution with mean µ = 205 pounds and standard deviation σ = 15 pounds.
Based on this information, what is the probability that all 49 boxes can be
safely loaded onto the freight elevator and transported?
Solution: Let Xk , for k = 1, 2, 3, . . . , 49, denote the weight of each
of the boxes. We want to estimate
!
n
X
Pr
Xk 6 9800 ,
k=1
√
where µ = E(Xk ) = 205 pounds, σ = VarXk = 15 pounds, and n =
49. We can therefore apply the Central Limit Theorem to estimate
!
Pn
n
X
9800 − 10, 045
k=1 Xk − nµ
√
Pr
Xk 6 9800
= Pr
6
7(15)
nσ
k=1
≈ Pr(Z 6 −2.33),
where Z ∼ Normal(0, 1), by the Central Limit Theorem. Therefore,
using the symmetry of the pdf of ∼ Normal(0, 1),
!
n
X
Pr
Xk 6 9800 ≈ 1 − FZ (2.33) =
˙ 1 − 0.9901 = 0.0099.
k=1
Thus, the probability that all 49 boxes can be safely loaded onto the
freight elevator and transported is about 0.99%, or less than 1%. 4. Forty–nine measurements are recorded to several decimal places. Each of these
49 numbers is rounded off to the nearest integer. The sum of the original
49 numbers is approximated by the sum of those integers. Assume that the
errors made in rounding off are independent, identically distributed random
variables with a uniform distribution over the interval (−0.5, 0.5). Compute
approximately the probability that the sum of the integers is within two units
of the true sum.
Math 151. Rumbos
Spring 2014
Solution: Let X1 , X2 , . . . , Xn , where n = 49, denote the 49 measurements, and Y1 , Y2 , . . . , Yn be the corresponding nearest integers after
rounding off. Then
Xi = Yi + Ui
for i = 1, 2, . . . , n,
where U1 , U2 , . . . , Un are independent identically distributed uniform
random variables on the interval (−0.5, 0.5). We then have that
E(Ui ) = 0 for all i, and
Var(Ui ) =
1
,
12
√
for all i. Thus, σ = 1/ 12.
The sum, S, of the original measurements is
S=
n
X
Yi +
i=1
where
n
X
n
X
Ui ,
i=1
Yi is the sum of the integer approximations. Put
i=1
W =S−
n
X
i=1
Yi =
n
X
Ui .
i=1
We would like to estimate Pr(|W | 6 2). We will do this by applying
the Central Limit Theorem to U1 , U2 , U3 , . . .
Observe that W = nU n , where U n is the sample mean of the rounding
off values U1 , U2 , . . . , Un . By the Central Limit Theorem
Un
√ 6 z ≈ Pr(Z 6 z),
Pr
σ/ n
for all z ∈ R, where Z ∼ Normal(0, 1). We therefore get that
|U n |
√ 6 z ≈ Pr(|Z| 6 z),
Pr
σ/ n
for z > 0.
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Math 151. Rumbos
Spring 2014
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It then follows that
Pr(|W | 6 2) = Pr(|U n | 6 2/n)
= Pr
|U n |
2
√ 6 √
σ/ n
σ n
2
≈ Pr |Z| 6 √
σ n
= 2FZ
= 2FZ
2
√
σ n
−1
√ !
4 3
√
−1
49
≈ 2FZ (0.99) − 1
≈ 2(0.8389) − 1
≈ 0.6778,
or about 67.78%. Thus, the probability that the sum of the 49 integers
is within 2 units of the true sum is about 67.78%.
5. Let X denote a random variable with pdf

 1 if 1 < x < ∞,
fX (x) = x2
0
otherwise.
Consider a random sample of size 72 from this distribution. Compute approximately the probability that 50 or more observations of the random sample are
less than 3.
Solution: The probability that a random observation from the distribution is less than 3 is given by
Z 3
Z 3
1
2
p=
fX (x) dx =
dx = .
2
3
−∞
1 x
Math 151. Rumbos
Spring 2014
Let Y denote the number of observation out of the 72 which are less
than 3. Then, Y ∼ Binomial(p, n), where n = 72. It then follows
that the
pmean of Y is µ = np = 48 and the standard deviation of Y
is σ = np(1 − p) = 4. By the Central Limit Theorem we then get
that
!
Y − np
6 z ≈ Pr(Z 6 z),
Pr p
np(1 − p)
for all z ∈ R, where Z ∼ Normal(0, 1), or
Y − 48
Pr
6 z ≈ Pr(Z 6 z).
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We want to estimate Pr(Y > 49) = 1 − Pr(Y 6 49), where
Y − 48
49 − 48
Pr(Y 6 49) = Pr
6
4
4
≈ Pr (Z 6 0.25)
≈ 0.5987.
Thus, Pr(Y > 49) ≈ 1 − 0.5987 = 0.4013 or about 40.13%.
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