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Statistics for Engineers and Scientists (4th Edition)
Chapter 2.1, Problem 4E Bookmark Show all steps: ON
Problem
Three times each day, a quality engineer samples a component from a recently manufactured batch and tests it. Each part is classified as conforming (suitable for its intended use), downgraded (unsuitable for the intended purpose but usable for another purpose), or scrap (not usable). An experiment consists of recording the categories of the three parts tested in a particular day.
a. List the 27 outcomes in the sample space.
b. Let A be the event that all the parts fall into the same category. List the outcomes in A.
c. Let B be the event that there is one part in each category. List the outcomes in B.
d. Let C be the event that at least two parts are conforming. List the outcomes in C.
e. List the outcomes in A ∩ C.
f. List the outcomes in A ∪ B.
g. List the outcomes in A ∩ Cc.
h. List the outcomes in Ac ∩ C.
i. Are events A and C mutually exclusive? Explain.
j. Are events B and C mutually exclusive? Explain.
Stepbystep solution
Let conforming is denoted by C, downgraded is denoted by D, and scrap is denoted by S.
Comment
(a)
The 27 outcomes in the sample space are shown below.
Step 1 of 11
Step 2 of 11
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S: {CCC, CCD, CCS, CDC, CDD, CDS, CSC, CSD, CSS,
DCC, DCD, DCS, DDC, DDD, DDS, DSC, DSD, DSS,
SCC, SCD, SCS, SDC, SDD, SDS, SSC, SSD, SSS}
Comment
(b)
Let A be the event that all the parts fall into the same category. The outcomes in A are shown below.
A: {CCC, DDD, SSS}
Comment
(c)
Let B be the event that there is one part in each category. The outcomes in B are shown below.
B: { CDS, CSD, DCS, DSC, SCD, SDC}
Comment
(d)
Let C be the event that at least two parts are conforming. The outcomes in C are shown below.
C: { CCC, CCD, CCS, CDC, CSC, DCC, SCC}
Comment
(e)
The sign denotes intersection. The event would be the event that all the parts fall into the same category and at least two parts are conforming.
: {CCC}
Comment
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(f)
The sign denotes union. The event would be the event that all the parts fall into the same category or there is one part in each category.
: {CCC, DDD, SSS, CDS, CSD, DCS, DSC, SCD, SDC}
Comment
(g)
The sign denotes complement. The event would contain the outcomes that are not in C.
: {CDD, CDS, CSD, CSS, DCD, DCS, DDC, DDD, DDS, DSC,
DSD, DSS, SCD, SCS, SDC, SDD, SDS, SSC, SSD, SSS}
The event would be the event that would contain the outcomes that are common in A and in .
: { DDD , SSS}
Comment
(h)
The sign denotes complement. The event would contain the outcomes that are not in A.
: {CCD, CCS, CDC, CDD, CDS, CSC, CSD, CSS,
DCC, DCD, DCS, DDC, DDS, DSC, DSD, DSS,
SCC, SCD, SCS, SDC, SDD, SDS, SSC, SSD}
The event would be the event that would contain the outcomes that are common in and in C.
: {CCD, CCS, CDC, CSC, DCC, SCC}
Comment
(i)
Two events are called mutually exclusive if they do not have any outcome in common. Events A and C are not mutually exclusive because they have common outcome CCC.
Comment
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(j)
Two events are called mutually exclusive if they do not have any outcome in common. Events B and C are mutually exclusive because they do not have any common outcome.
Comment
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Statistics for Engineers and Scientists (4th Edition)
Chapter 2.3, Problem 8E Bookmark Show all steps: ON
Problem
A drag racer has two parachutes, a main and a backup, that are designed to bring the vehicle to a stop after the end of a run. Suppose that the main chute deploys with probability 0.99, and that if the main fails to deploy, the backup deploys with probability 0.98.
a. What is the probability that one of the two parachutes deploys?
b. What is the probability that the backup parachutes deploys?
Stepbystep solution
(a) Let A denote the event that a main parachute deploys and let B denote the event that a backup parachute deploys. We are given the following probabilities:
Comment
The event that one of the parachutes deploys is a complement to the event that none of the parachutes deploy. Find .
Comment
(b) By the law of total probability,
If a main parachute deploys, the probability that a backup parachute deploys is equal to 0, that is . Thus,
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Statistics for Engineers and Scientists (4th Edition)
Chapter 2.3, Problem 24E Bookmark Show all steps: ON
Problem
A lot of 1000 components contains 300 that are defective. Two components are drawn at random and tested. Let A be the event that the first component drawn is defective, and let B be the event that the second component drawn is defective.
a. Find P(A).
b. Find P(B|A) .
c. Find P(A ∩ B).
d. Find P(Ac ∩ B).
e. Find P(B) .
f. Find P(A|B).
g. Are Aand B independent? Is it reasonable to treat A and B as though they were independent? Explain.
Stepbystep solution
24. (a) P(A) = 300/1000 = 3/10
Comment
(b) Given that A occurs, there are 999 components remaining, of which 299 are defective. Therefore P(B|A) = 299/999.
Comment
(c)
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Comment
(d) Given that Ac occurs, there are 999 components remaining, of which 300 are defective.
Therefore
Comments (1)
(e)
Comment
(f)
Comment
(g) A and B are not independent, but they are very nearly independent. To see this note that P(B) = 0.3, while P(B|A) = 0.2993. So P(B) is very nearly equal to P(B|A), but not exactly equal. Alternatively, note that Therefore in most situations it would be reasonable to treat A and B as though they were independent.
Comment
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Statistics for Engineers and Scientists (4th Edition)
Chapter 2.4, Problem 2E Bookmark Show all steps: ON
Problem
Computer chips often contain surface imperfections. For a certain type of computer chip, the probability mass function of the number of defects X is presented in the following table.
a. Find P(X ≤ 2).
s
b. Find P(X > 1).
c. Find μX.
d. Find
Stepbystep solution
(a) To compute we must sum the probabilities for the values of that are less
than or equal to 2, namely 0, 1, and 2. Thus,
Comment
(b) To compute we must sum the probabilities for the values of that are greater
than 1, namely 2, 3 and 4. Thus,
Comment
Step 1 of 4
Step 2 of 4
Chapter 2.4
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(c) Using the equation , compute the mean of X.
Comment
(d) Compute the variance by using an alternate formula .
Comment
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Statistics for Engineers and Scientists (4th Edition)
Chapter 2.4, Problem 4E Bookmark Show all steps: ON
Problem
Let X represent the number of tires with low air pressure on a randomly chosen car.
a. Which of the three functions below is a possible probability mass function of X? Explain.
X
0 1 2 3 4
p1(x) 0.2 0.2 0.3 0.1 0.1
p2(x) 0.1 0.3 0.3 0.3 0.2
p3(x) 0.1 0.2 0.2 0.4 0.1
b. For the possible probability mass function, compute μX. and
Stepbystep solution
(a) If X is a random variable representing the number of tires with low air pressure on a
randomly chosen car, then , where the sum is over all the possible values of
X. To determine which of the three functions given could be a possible probability mass
function of X we compute the sum for each , where i is from 1 to 3.
Comment
For the first function we get:
Since , cannot be a probability mass function of X.
Comment
Step 1 of 6
Step 2 of 6
Chapter 2.4
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For the second function we get:
Since , cannot be a probability mass function of X.
Comment
For the third function we get:
As far as , is a possible probability mass function of X.
Comment
(b) Using the equation compute the mean of X.
Comment
Compute the variance by using the formula .
Comment
Step 3 of 6
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Step 5 of 6
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Statistics for Engineers and Scientists (4th Edition)
Chapter 2.4, Problem 12E Bookmark Show all steps: ON
Problem
Three components are randomly sampled, one at a time, from a large lot. As each component is selected, it is tested. If it passes the test, a success (S) occurs; if it fails the test, a failure (F) occurs. Assume that 80% of the components in the lot will succeed in passing the test. Let X represent the number of successes among the three sampled components.
a. What are the possible values for X?
b. Find P(X = 3).
c. The event that the first component fails and the next two succeed is denoted by FSS. Find P(FSS)
d. FindP(SFS) and P(SSF).
e. Use the results of parts (c) and (d) to find P(X = 2).
f. Find P(X= 1).
g. Find P(X = 0).
h. Find μ x
i. Find
j. Let Y represent the number of successes if four components are sampled. Find P(Y = 3).
Stepbystep solution
12. (a) 0, 1, 2, 3
Comment
(b) P(X = 3) = P(SSS) = (0.8)3 = 0.512
Step 1 of 10
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Comments (1)
(c) P(FSS) = (0.2)(0.8)2 = 0.128
Comment
(d) P(SFS) = P(SSF) = (0.8)2(0.2) = 0.128
Comment
(e) P(X = 2) = P(FSS) + P(SFS) + P(SSF) = 0.384
Comment
(f)
Comment
(g) P(X = 0) = P(FFF) = (0.2)3 = 0.008
Comment
(h) μX = 0(0.08) + 1(0.096) + 2(0.384) + 3(0.512) = 2.4
Comment
(i)
Alternatively,
Comment
Step 3 of 10
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(j)
Comment
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Statistics for Engineers and Scientists (4th Edition)
Chapter 4.2, Problem 2E Bookmark Show all steps: ON
Problem
Let X ~ Bin(9, 0.4). Find
a. P(X > 6)
b. P(X ≥ 2)
c. P(2 ≤ X < 5)
d. P(2 < X ≤ 5)
e. P(X = 0)
f. P(X = 7)
g. μx
h.
Stepbystep solution
(a)
Let . Therefore, and .
Comments (1)
(b)
Let . Therefore, and .
Step 1 of 8
Step 2 of 8
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Comments (2)
(c)
Let . Therefore, and .
Comment
(d)
Let . Therefore, and .
Comment
(e)
Let . Therefore, and .
Comment
(f)
Let . Therefore, and .
Step 3 of 8
Step 4 of 8
Step 5 of 8
Step 6 of 8
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Comment
(g)
Let . Therefore, and . Therefore, the mean is given as:
Comment
(f)
Let . Therefore, and . Therefore, the variance is given as:
Comment
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Statistics for Engineers and Scientists (4th Edition)
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Chapter 4.2, Problem 10E Bookmark Show all steps: ON
Problem
A quality engineer takes a random sample of 100 steel rods from a day's production, and finds that 92 of them meet specifications.
a. Estimate the proportion of that day's production that meets specifications, and find the uncertainty in the estimate.
b. Estimate the number of rods that must be sampled to reduce the uncertainty to 1%.
Stepbystep solution
10. Let X be the number of rods in the sample that meet specifications, and let p be the proportion of the day's production that meets specifications. Then X ~ Bin(100, p). The observed value of X is 92.
(a)
Replacing p with and n with 100,
Comment
(b) We must find n so that
Substitute for p to obtain Solving for n yields n = 736.
Comment
Step 1 of 2
Step 2 of 2
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Statistics for Engineers and Scientists (4th Edition)
Chapter 4.3, Problem 5E Bookmark Show all steps: ON
Problem
A data center contains 1000 computer servers. Each server has probability 0.003 of failing on a given day.
a. What is the probability that exactly two servers fail?
b. What is the probability that fewer than 998 servers function?
c. What is the mean number of servers that fail?
d. What is the standard deviation of the number of servers that fail?
Stepbystep solution
(a)
The probability of failing a server on a given day is and there are total computer servers. The mean number of fail per day is:
Let X denotes number of servers fail.
Here, the random variable X follows Poisson distribution with parameter
The probability function for the Poisson random variable is,
Comment
Compute the probability that exactly two servers fail on a day.
The probability that exactly two servers fail on a day is
Step 1 of 5
Step 2 of 5
Chapter 4.3
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Comment
(b)
The probability that fewer than 998 servers functions means more than 2 servers fail on a day, that is , is given as:
The probability that fewer than 998 servers functions is
Comment
(c)
The probability of failing a server on a given day is and there are total computer servers. The mean number of servers fail per day is:
The mean number of servers fail per day is
Comment
(d)
Compute the standard deviation of number of fail per day.
The standard deviation of number of fail per day is
Comment
Step 3 of 5
Step 4 of 5
Step 5 of 5
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Chapter 4.3, Problem 8E Bookmark Show all steps: ON
Problem
The number of cars arriving at a given intersection follows a Poisson distribution with a mean rate of 4 per second.
a. What is the probability that 3 cars arrive in a given second?
b. What is the probability that 8 cars arrive in three seconds?
c. What is the probability that more than 3 cars arrive in a period of two seconds?
Stepbystep solution
(a)
Let . The probability that 3 cars arrive in a given second, that is , is
given as:
This is because Poisson Probability Distribution:
Where
Comment
(b)
Let . The probability that 8 cars arrive in three seconds, that is and
, is given as:
This is because Poisson Probability Distribution:
Where
Step 1 of 3
Step 2 of 3
Chapter 4.3
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Comment
(c)
Let . The probability that more than 3 cars arrive in two seconds, that is
and , is given as:
Comment
Step 3 of 3
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