Stat 330 (Spring 2015): Homework 7
Due: March 13th, 2015
Show all of your work, and please staple your assignment if you use more than one sheet. Write your name,
the course number and the section on every sheet. Problems marked with * will be graded and one additional
randomly chosen problem will be graded. Must show all work for full credit.
1. * The price of a particular make of a 64GB iPad Mini among dealers nationwide is assumed to have a
Normal distribution with mean µ = $500 with standard deviation σ = $20.
(a.) What is the probability that an iPad Mini of the same specs, chosen randomly from a dealer, will
cost less than $460? (Show your work).
Answer:
460 − 500
P (X < 460) = P Z <
20
= P (Z < −2) = Φ(−2) = 1 − Φ(2) = 1 − 0.9772 = 0.0228.
(b.) What is the probability that an iPad Mini of the same specs, chosen randomly from a dealer, will
cost greater than $549? (Show your work).
Answer:
549 − 500
P (X > 549) = P Z >
20
= P (Z > 2.45) = 1 − Φ(2.45) = 1 − 0.9929 = 0.0071.
(c.) What is the probability that an iPad Mini of the same specs, chosen randomly from a dealer, will
cost between $525 and $545? (Show your work).
Answer:
545 − 500
525 − 500
<Z<
P (525 < X < 545) = P
20
20
= P (1.25 < Z < 2.25)
=
Φ(2.25) − Φ(1.25) = 0.9878 − 0.8944 = 0.0934.
(d.) The manufacturer doesn’t want dealers to markup these iPad Minis above the 90th percentile of the
price distribution. Approximately, what is the 90th percentile of the distribution of the price of this
iPad Mini? Show your work.
Answer:
P (Z < z) = 0.90 =⇒ z = 1.28
Hence we have
X − 500
= 1.28 =⇒ X − 500 = 25.6 =⇒ X = $525.60.
20
(e.) What is the probability that the average price of 25 iPad Minis of the same specs, chosen randomly
from a dealer, will be greater than $510? (Show your work).
Answer: The average price of 25 computers
make chosen randomly from a dealer has
√
√ of the same
mean µ = $500 and standard deviation σ/ n = 20/ 25= $4.
Hence the probability that the average price of 25 computers of the same make is greater than
$510 is
510 − 500
P (X̄ > 510) = P Z >
4
= P (Z > 2.5) = 1 − Φ(2.5) = 1 − 0.9938 = 0.0062.
1
Stat 330 (Spring 2015): Homework 7
Due: March 13th, 2015
(f.) Compute the probability that at least 4 out of 5 iPad Minis of the same specs picked randomly from
among dealers will cost between $525 and $545?
Answer:
Let Y denote r.v. that represent the number of computers out of 5 that will cost
between$525 and $545. Thus, denoting a “Success” as the event “a computer will cost between $525
and $545 ”, we have that
Y ∼ Bin(5, p)
where p = P (“a computer will cost between $525 and $545”) = P (525 < X < 545) = .0934 from
part (c).
The required probabilty is thus
P (Y = 4) + P (Y = 5)
5
5
(.0934)4 (1 − .0934)1 +
(.0934)5 (1 − .0934)0
4
5
= 5(.0000761).9066 + 7.10779e − 6
=
= .0003520678
2. Consider two brands of batteries for calculators, Xmax and Y-cell, manufactured independently by two
companies. Assume that the lifetimes of each battery has a Normal distribution. The Xmax batteries
has a mean of 1000 hours with a standard deviation of 100 hours, and Y-cell batteries has a mean of
1200 hours with a standard deviation of 200 hours. For each of the following questions, you must state
the random variable you are using and the distribution assumption you make.
(a.) What is the probability of an Xmax battery will last at most 800 hours? What is this probability
for a Y-cell battery?
(b.) What is the probability that the average lifetime of 400 Xmax batteries is less than 1012 hours?
(c.) Assume, two batteries are randomly selected, one Xmax and one Y-cell. What can you say about
the difference D in their lifetimes? That is, find the distribution of D, it’s expected value and
variance.
(d.) You notice that, on average, Y-cell batteries will last longer than Xmax batteries. But what is the
exact probability that the Y-cell batteries will last longer than Xmax batteries?
Answer:
Let X = the lifetime of Xmax batteries
Y = the lifetime of Y-cell batteries.
X ∼ N(1000, 1002 )
Y ∼ N(1200, 2002 )
(a)
P [X ≤ 800]
800 − 1000
= P Z≤
100
= P [Z ≤ −2]
= P [Z ≥ 2] = 0.0228
(from Tables)
800 − 1200
P [Y ≤ 800] = P Z ≤
200
= P [Z ≤ −2] = 0.0228
(b) X̄ = mean lifetime of 400 Xmax batteries.
1002
X̄ ∼ N 1000,
400
2
Stat 330 (Spring 2015): Homework 7
P [X < 1012]
Due: March 13th, 2015
=
1012 − 1000
P Z<
100/20
P [Z < 12/5]
=
P [Z < 2.4] = 0.9918
=
(c) Take D = Y − X (could take D = X − Y )
This is of the form aX + bY so distributed Normal.
E[D] = E[Y ] − E[X] = 1200 − 1000 = 200
V ar[D] = V ar[Y ] + V ar[X]
2
(X and Y are independent r.v.’s)
2
= 200 + 100 = 50, 000
Thus D ∼ N(200, 50000)
(d) Need
P [(Y − X) > 0](or
0 − 200
=P Z > √
50000
=P [Z > −0.8944]
P [(X − Y ) > 0])
=1 − P [Z ≤ −0.8944] = 0.8144
3. An autopilot attempts to maintain an airplane’s altitude at a constant 18000 feet. Because of random
variations, the actual altitude X is Normal random variable with parameters (µ = 18000, σ 2 = 2500).
(a) What is the probability that the altitude differs from the assigned 18000 value by more than 100
feet?
(b) Suppose that by mechanically improving the autopilot, the variation of the airplane’s altitude σ 2
can actually be reduced. Find σ 2 such that the probability that the altitude differs from the assigned
18000 value by more than 100 feet is at most 0.01.
Answer:
(a) Given X ∼ N (18, 000, 2500), we need to calculate P (|X − 18000| > 100):
P (|X − 18000| > 100)
= P (X < 17, 900 or X > 18, 100)
1 − P (17, 900 < X < 18, 100)
17, 900 − 18, 000
18, 100 − 18, 000
= 1−P
<Z<
50
50
= 1 − P (−2 < Z < 2) = 1 − (Φ(2) − Φ(−2))
=
=
1 − (0.9772 − 0.0228) = 1 − 0.9544 = 0.0456
(b) Need to find σ so that
P (|X − 18000| > 100) ≤ 0.01
P
P (|X − 18000| ≤ 100) > 0.99
17, 900 − 18, 000
18, 100 − 18, 000
> 0.99
<Z<
σ
σ
or, equivalently
P
−100
100
<Z<
σ
σ
> 0.99
implying that we need 100/σ > 2.575, noting that Φ(2.575) − Φ(−2.575) = 0.99. This gives
σ < 38.835, which implies σ 2 < 1508.16.
3
Stat 330 (Spring 2015): Homework 7
Due: March 13th, 2015
4. * (Baron’s book): 4.22
Answer:
36
Instructor’s solution manual
5. (Baron’s Continuity
book): 4.25 correction is not needed, because Xi , the times, are already continuous
variables.
Answer:
4.25
(a) The number X of defective chips in the sample has Binomial distribution with
parameters n p
= 400 and p = 0.06, expectation µ = np = 24, and standard
deviation σ = np(1 − p) = 4.75.
By the Central Limit Theorem (with a suitable continuity correction),
P {20 ≤ X ≤ 25} = P {19.5 ≤ X ≤ 25.5}
19.5 − 24
25.5 − 24
= P
≤Z ≤
= P {−0.95 ≤ Z ≤ 0.32}
4.75
4.75
≈ Φ(0.32) − Φ(−0.95) = 0.6255 − 0.1711 = 0.4544
from Table A4.
(b) The number Y of inspectors that find between 20 and 25 defective chips in their
samples has Binomial distribution with parameters n = 40 and p = 0.4544,
computed
in (a), expectation µ = np = 18.176, and standard deviation σ =
p
np(1 − p) = 3.149. Then
7.5 − 18.176
P {X ≥ 8} = P {X ≥ 7.5} = P Z ≥
3.149
= P {Z ≥ −3.39} ≈ Φ(3.39) = 0.9997
from Table A4.
4.26 Here n = 80, µ = 0.6, and σ = 0.4.
50 − nµ
nµ
6. (Baron’s book): 4.31 P {47 ≤ S80 ≤ 50} = P 47 −
√
√
≤Z ≤
σ n
σ n
Answer:
47 − 48
50 − 48
√ ≤Z≤
√
= P
≈ Φ(0.56) − Φ(−0.28)
0.4 80
0.4 80
= 0.7123 − 0.3897 = 0.3226
from Table A4.
4.27 The number of damaged
p files has Binomial distribution with n = 2400, p = 0.35,
µ = np = 840, and σ = np(1 − p) = 23.37. Then
P {800 ≤ X ≤ 850} = P {799.5 < X < 850.5}
4850.5 − 840 799.5 − 840
= P
= P {−1.73 < Z < 0.45}
<Z<
23.37
23.37
≈
Φ(0.45) − Φ(−1.73) = 0.6736 − 0.0418 = 0.6318
Stat 330 (Spring 2015): Homework 7
Due: March 13th, 2015
This question and the answer above are a bit incomplete. The question should really state that at least
95% of the time upgrading takes less than 10 minutes. Thus, the probability statement is
0.95 ≤ P (SN < 600 sec).
The roots of 225N 2 −18030N +360000 = 0 are 37.756 and 42.377 and thus there are really 4 possibilities:
37, 38, 42, and 43. The question is which one of these satisfies the probability statement above. It turns
out that only 37 does since
P (S37 < 600) ≈ 0.987
and
So the final answer should be N = 37.
5
P (S38 < 600) ≈ 0.929.