/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q69E Each of \({\rm{12}}\)refrigerato... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Q69E (page 130)

Each of \({\rm{12}}\)refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerators are running. Suppose that \({\rm{7}}\) of these refrigerators have a defective compressor and the other \({\rm{5}}\) have less serious problems. If the refrigerators are examined in random order, let\({\rm{X}}\)be the number among the first \({\rm{6}}\) examined that have a defective compressor.

a. Calculate\({\rm{P(X = 4)}}\)and \(P(X拢 4)\)

b. Determine the probability that \({\rm{X}}\) exceeds its mean value by more than \({\rm{1}}\) standard deviation.

c. Consider a large shipment of \({\rm{400}}\)\({\rm{40}}\) refrigerators, of which 40 have defective compressors. If \({\rm{X}}\) is the number among \({\rm{15}}\) randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) P(X拢5)than to use the hypergeometric \({\rm{pmf}}\).

Short Answer

Expert verified

a). \(\begin{array}{*{20}{c}}{{\rm{P(X = 4) = }}\frac{{{\rm{25}}}}{{{\rm{66}}}}{\rm{\gg 0}}{\rm{.3788 = 37}}{\rm{.88\% }}}\\{{\rm{P(X拢4) = }}\frac{{{\rm{29}}}}{{{\rm{33}}}}{\rm{\gg 0}}{\rm{.8788 = 87}}{\rm{.88\% }}}\end{array}\)

b). \({\rm{P(X > \mu + \sigma ) = }}\frac{{\rm{4}}}{{{\rm{33}}}}{\rm{\gg 0}}{\rm{.1212 = 12}}{\rm{.12\% }}\)

c). binominal distribution: \({\rm{n = 15}}\)and \({\rm{p = 0}}{\rm{.1}}\)

\({\rm{P(X拢5)\gg 0}}{\rm{.9978 = 99}}{\rm{.78\% }}\)

Step by step solution

01

Definition of probability

The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

 Calculating \({\rm{P(X = 4) and P(X£4)}}\)

Given: The number of refrigerators with faulty compressors is \({\rm{X}}\)

\(\begin{array}{*{20}{c}}{}&{{\rm{N = Population size = 12}}}\\{}&{{\rm{n = Number of draws = 6}}}\end{array}\)

\({\rm{r = }}\)Number of successes observed \({\rm{ = 7}}\)

Hypergeometric probability formula:

\({\rm{p(y) = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{r}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{\rm{N - r}}}\\{{\rm{n - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{l}}{\rm{N}}\\{\rm{n}}\end{array}} \right)}}\)

Calculate at \({\rm{y = 1,2,3,4}}\) (There are \({\rm{7}}\)refrigerators with defective compressors out of \({\rm{12}}\) refrigerators, hence there can only be \({\rm{5}}\)refrigerators with no defective compressors and at least one defective compressor in the sample of \({\rm{6}}\))

\({\rm{p(0) = 0(0refrigerators with a defective compressot}}\)

\(\begin{array}{*{20}{c}}{}\\{{\rm{p(1) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{1}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 1}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{132}}}}}\end{array}\)

\(\begin{array}{*{20}{c}}{}\\{{\rm{p(2) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 2}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{\rm{5}}}{{{\rm{44}}}}}\end{array}\)

\(\begin{array}{*{20}{c}}{}\\{{\rm{p(3) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{3}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 3}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{{\rm{25}}}}{{{\rm{66}}}}}\end{array}\)

\({\rm{P(X = 4) = p(4) = }}\frac{{{\rm{(4)(0 - 4)}}}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{{\rm{2b}}}}{{{\rm{66}}}}{\rm{\gg 0}}{\rm{.3788 = 37}}{\rm{.88\% }}\)

For discontinuous or mutually exclusive occurrences, use the following addition rule:

\({\rm{P(A\;or\;B) = P(A) + P(B)}}\)

Compile the following probabilities:

\(\begin{array}{*{20}{c}}{{\rm{P(X拢4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)}}}\\{{\rm{ = 0 + }}\frac{{\rm{1}}}{{{\rm{132}}}}{\rm{ + }}\frac{{\rm{5}}}{{{\rm{44}}}}{\rm{ + }}\frac{{{\rm{25}}}}{{{\rm{66}}}}{\rm{ + }}\frac{{{\rm{25}}}}{{{\rm{66}}}}{\rm{ = }}\frac{{{\rm{116}}}}{{{\rm{132}}}}{\rm{ = }}\frac{{{\rm{29}}}}{{{\rm{33}}}}{\rm{\gg 0}}{\rm{.8788 = 87}}{\rm{.88\% }}}\end{array}\)

03

Determining the probability that \({\rm{X}}\) exceeds its mean value by more than \({\rm{1}}\) standard deviation

Given: The number of refrigerators with faulty compressors is \({\rm{X}}\)

\(\begin{array}{*{20}{c}}{}&{{\rm{N = Population size = 12}}}\\{}&{{\rm{n = Number of draws = 6}}}\end{array}\)

\({\rm{r = }}\)Number of successes observed \({\rm{ = 7}}\)

The product of the number of draws and the number of observed successes, divided by the population size, is the mean of a hypergeometric distribution:

\({\rm{\mu = }}\frac{{{\rm{n \times r}}}}{{\rm{N}}}{\rm{ = }}\frac{{{\rm{6 \times 7}}}}{{{\rm{12}}}}{\rm{ = }}\frac{{{\rm{42}}}}{{{\rm{12}}}}{\rm{ = }}\frac{{\rm{7}}}{{\rm{2}}}{\rm{ = 3}}{\rm{.5}}\)

The hypergeometric distribution's variance is:

\({{\rm{\sigma }}^{\rm{2}}}{\rm{ = }}\frac{{{\rm{N - n}}}}{{{\rm{N - 1}}}}{\rm{ \times n \times }}\frac{{\rm{r}}}{{\rm{N}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{r}}}{{\rm{N}}}} \right){\rm{ = }}\frac{{{\rm{12 - 6}}}}{{{\rm{12 - 1}}}}{\rm{ \times 6 \times }}\frac{{\rm{7}}}{{{\rm{12}}}}{\rm{ \times }}\left( {{\rm{1 - }}\frac{{\rm{7}}}{{{\rm{12}}}}} \right){\rm{ = }}\frac{{{\rm{35}}}}{{{\rm{44}}}}{\rm{\gg 0}}{\rm{.7955}}\)

Find the value that is one standard deviation higher than the mean. The square root of the variance is the standard deviation:

\({\rm{\mu + \sigma = 3}}{\rm{.5 + }}\sqrt {{\rm{0}}{\rm{.7955}}} {\rm{\gg 4}}{\rm{.3919}}\)

Hypergeometric probability formula:

\({\rm{p(y) = }}\frac{{\left( {\begin{array}{*{20}{l}}{\rm{r}}\\{\rm{y}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{\rm{N - r}}}\\{{\rm{n - y}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{l}}{\rm{N}}\\{\rm{n}}\end{array}} \right)}}\)

If \({\rm{X}}\)is equal to \({\rm{5}}\) or\({\rm{6}}\) (\({\rm{X}}\)cannot take on values bigger than \({\rm{6}}\)), it is larger than \({\rm{\mu + \sigma = 4}}{\rm{.3919}}\)

\(\begin{array}{*{20}{c}}{{\rm{p(5) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{5}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 5}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{\rm{5}}}{{{\rm{44}}}}}\\{{\rm{p(6) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{7}}\\{\rm{6}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{12 - 7}}}\\{{\rm{6 - 6}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{12}}}\\{\rm{6}}\end{array}} \right)}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{132}}}}}\end{array}\)

For discontinuous or mutually exclusive occurrences, use the following addition rule:

\({\rm{P(A\;or\;B) = P(A) + P(B)}}\)

Compile the following probabilities:

\(\begin{array}{*{20}{c}}{{\rm{P(X > \mu + \sigma ) = P(X > 4}}{\rm{.3919) = P(X = 5) + P(X = 6)}}}\\{{\rm{ = }}\frac{{\rm{5}}}{{{\rm{44}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{132}}}}{\rm{ = }}\frac{{{\rm{16}}}}{{{\rm{132}}}}{\rm{ = }}\frac{{\rm{4}}}{{{\rm{33}}}}{\rm{\gg 0}}{\rm{.1212 = 12}}{\rm{.12\% }}}\end{array}\)

04

Determining the a less tedious way to calculate (at least approximately) \({\rm{P(X\pounds5)}}\) than to use the hypergeometric \({\rm{pmf}}\).

Given: The number of refrigerators with faulty compressors is \({\rm{X}}\)

\(\begin{array}{*{20}{c}}{}&{{\rm{N = Population size = 400}}}\\{}&{{\rm{n = Number of draws = 12}}}\end{array}\)

\({\rm{r = }}\)Number of successes observed \({\rm{ = 40}}\)

When the population size is big, the binomial distribution with \({\rm{n = 15}}\)is used to approximate the hypergeometric distribution.

\({\rm{p = }}\frac{{\rm{r}}}{{\rm{N}}}{\rm{ = }}\frac{{{\rm{40}}}}{{{\rm{400}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{ = 0}}{\rm{.1}}\)

Binomial probability definition:

\({\rm{P(X = k)}}{{\rm{ = }}_{\rm{n}}}{{\rm{C}}_{\rm{k}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}\)

Examine at\({\rm{k = 0,1,2,3,4,5}}\):

\({\rm{P(X = 0) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 0)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{0}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 0}}}}{\rm{\gg 0}}{\rm{.2059}}\)

\(\begin{array}{*{20}{c}}{{\rm{P(X = 1) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 1)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{1}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 1}}}}{\rm{\gg 0}}{\rm{.3432}}}\\{{\rm{P(X = 2) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 2)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{2}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 2}}}}{\rm{\gg 0}}{\rm{.2669}}}\\{{\rm{P(X = 3) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 3)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{3}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 3}}}}{\rm{\gg 0}}{\rm{.1285}}}\\{{\rm{P(X = 4) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 4)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{4}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 4}}}}{\rm{\gg 0}}{\rm{.0428}}}\end{array}\)

\({\rm{P(X = 5) = }}\frac{{{\rm{15!}}}}{{{\rm{15!(15 - 5)!}}}}{\rm{ \times 0}}{\rm{.}}{{\rm{1}}^{\rm{5}}}{\rm{ \times (1 - 0}}{\rm{.1}}{{\rm{)}}^{{\rm{15 - 5}}}}{\rm{\gg 0}}{\rm{.0105}}\)

For discontinuous or mutually exclusive occurrences, use the following addition rule:

\({\rm{P(A\;or\;B) = P(A) + P(B)}}\)

Compile the following probabilities:

\(\begin{array}{*{20}{c}}{{\rm{P(X拢5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)}}}\\{{\rm{ = 0}}{\rm{.2059 + 0}}{\rm{.3432 + 0}}{\rm{.2669 + 0}}{\rm{.1285 + 0}}{\rm{.0428 + 0}}{\rm{.0105\gg 0}}{\rm{.9978 = 99}}{\rm{.78\% }}}\end{array}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

After shuffling a deck of \({\rm{52}}\) cards, a dealer deals out\({\rm{5}}\). Let \({\rm{X = }}\) the number of suits represented in the five-card hand.

a. Show that the pmf of \({\rm{X}}\) is

(Hint: \({\rm{p(1) = 4P}}\) (all are spades), \({\rm{p(2) = 6P}}\) (only spades and hearts with at least one of each suit), and \({\rm{p(4)}}\) \({\rm{ = 4P(2}}\) spades \({\rm{{C}}}\) one of each other suit).)

b. Compute\({\rm{\mu ,}}{{\rm{\sigma }}^{\rm{2}}}\), and\({\rm{\sigma }}\).

Show that E(X) = np when X is a binomial random variable. (Hint: First express E(X) as a sum with lower limit x =\({\rm{1}}\). Then factor out np, let y = x\({\rm{ - 1}}\)so that the sum is from y = 0 to y = n\({\rm{ - 1}}\), and show that the sum equals\({\rm{1}}\).)

a. In a Poisson process, what has to happen in both the time interval \({\rm{(0,t)}}\) and the interval \({\rm{(t,t + \Delta t)}}\) so that no events occur in the entire interval \({\rm{(0,t + \Delta t)}}\)? Use this and Assumptions \({\rm{1 - 3}}\) to write a relationship between \({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t)}}\) and \({{\rm{P}}_{\rm{0}}}{\rm{(t)}}\)

b. Use the result of part (a) to write an expression for the difference \({{\rm{P}}_{\rm{0}}}{\rm{(t + \Delta t) - }}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}\) Then divide by \({\rm{\Delta t}}\) and let to obtain an equation involving \({\rm{(d/dt)}}{{\rm{P}}_{\rm{0}}}{\rm{(t)}}\), the derivative of \({{\rm{P}}_{\rm{0}}}{\rm{(t)}}\)with respect to \({\rm{t}}\).

c. Verify that \({{\rm{P}}_{\rm{0}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}\)satisfies the equation of part (b).

d. It can be shown in a manner similar to parts (a) and (b) that the \({{\rm{P}}_{\rm{k}}}{\rm{(t)}}\)must satisfy the system of differential equations

\(\begin{array}{*{20}{c}}{\frac{{\rm{d}}}{{{\rm{dt}}}}{{\rm{P}}_{\rm{k}}}{\rm{(t) = \alpha }}{{\rm{P}}_{{\rm{k - 1}}}}{\rm{(t) - \alpha }}{{\rm{P}}_{\rm{k}}}{\rm{(t)}}}\\{{\rm{k = 1,2,3, \ldots }}}\end{array}\)

Verify that \({{\rm{P}}_{\rm{k}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}{{\rm{(\alpha t)}}^{\rm{k}}}{\rm{/k}}\) satisfies the system. (This is actually the only solution.)

Two fair six-sided dice are tossed independently. Let M = the maximum of the two tosses (so M(1,5) =5, M(3,3) = 3, etc.).

a. What is the pmf of M? (Hint: First determine p(1), then p(2), and so on.)

b. Determine the cdf of M and graph it.

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives circuit boards in batches of five. Two boards are selected from each batch for inspection. We can represent possible outcomes

of the selection process by pairs. For example, the pair (1, 2) represents the selection of boards 1 and 2 for inspection.

a. List the ten different possible outcomes.

b. Suppose that boards 1 and 2 are the only defective boards in a batch. Two boards are to be chosen at random. Define X to be the number of defective boards observed among those inspected. Find the probability distribution of X.

c. Let F(x) denote the cdf of X. First determine \(F\left( 0 \right) = P\left( {X \le 0} \right)\), F(1), and F(2); then obtain F(x) for allother x.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.