Stat 330 (Spring 2015): Homework 4
Due: February 16, 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.
1. * When circuit boards used in the manufacture of portable disk drives are tested, the long-run percentage
of defectives is 5%. Assume that the events that the boards are defective or not occur independently of
each other.
Answer: Given that 5% of the circuit boards are defective, the probability of getting a defective
computer in a batch is 0.05.
Let us denote the random variable X to represent the number of defective circuit boards in a sample of
size 25. Assuming that the probability of getting a defective circuit board is the same for any circuit
board, X ∼ Binomial(n = 25, p = 0.05). We’ll use X to answer part (a) to (c)
(a) Calculate the probability of finding exactly 3 defective circuit boards in a batch of twenty five.
Answer:
25
P [X = 3] =
(0.05)3 (1 − 0.05)25−3 = 0.0930
3
(b) Calculate the probability of finding at most 3 defective circuit boards in a batch of twenty five.
Answer:
3 X
25
P [X ≤ 3] =
(0.05)x (1 − 0.05)25−x
x
x=0
= P [X = 0] + P [X = 1] + P [X = 2] + P [X = 3]
25
25
25
25
0
25
1
24
2
23
=
(0.05) (.95) +
(0.05) (.95) +
(0.05) (.95) +
(0.05)3 (.95)22
0
1
2
3
= 0.2774 + 0.3650 + 0.2305 + .0930 = 0.9659
(c) Calculate the probability of finding more than 1 but less than 4 defective circuit boards in a batch
of twenty five.
Answer: P [1 < X < 4] = P [X = 2] + P [X = 3] = 0.2305 + .0930 = 0.3235
(d) Calculate the probability that at least 5 circuit boards have to be tested before a defective circuit
board is found.
Answer: Let Y = number of circuit boards before he finds a defective circuit board. Note that
that each test of a circuit board is a Bernoulli trial with P (defective) = .05 the same for each trial.
Then Y ∼ Geometric(0.05) The probability that the engineer has to test at least 5 circuit boards
before he finds a defective circuit board is
P [Y ≥ 5] = 1 − P [Y ≤ 4] = 1 − F (4) = 1 − (1 − (.95)4 ) = 0.8145
since F (t) = P [Y ≤ t] = 1 − (1 − p)t is the cdf of the Geometric distribution.
2. A digital communication channel has an error rate of 10−7 , which may be taken to mean that the
probability of any bit (a one or a zero) arriving incorrectly at the destination is 10−7 . What is the
probability that a transmission of a million bits occurs without a single error? It is reasonable to assume
that the transmission of a given bit (incorrectly or correctly) does not affect the transmission of other
bits. State clearly the random variable and the probability distribution you are using and the values of
the parameters associated with that distribution.
Answer: Let X be the random variable representing the number of bits transmitted incorrectly out of
a million bits.
Then under the given conditions, X has a Binomial distribution with parameters n = 106 and p = 10−7 .
1
Stat 330 (Spring 2015): Homework 4
Due: February 16, 2015
The probability that a transmission of a million bits occurs without a single error is P (X = 0)
6
6
10
P (X = 0) =
· (10−7 )0 · (1 − 10−7 )10 = 0.905
0
3. (Baron’s book): 3.1
Answer:
4. * (Baron’s book): 3.18
Answer:
2
Stat 330 (Spring 2015): Homework 4
Due: February 16, 2015
5. (Baron’s book): 3.23
Answer:
3