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Chapter 5: Q41E (page 229)

Let \({\rm{X}}\) be the number of packages being mailed by a randomly selected customer at a certain shipping facility. Suppose the distribution of \({\rm{X}}\) is as follows:

a. Consider a random sample of size \({\rm{n = 2}}\) (two customers), and let \({\rm{\bar X}}\) be the sample mean number of packages shipped. Obtain the probability distribution of\({\rm{\bar X}}\).

b. Refer to part (a) and calculate\({\rm{P(\bar X\pounds2}}{\rm{.5)}}\).

c. Again consider a random sample of size\({\rm{n = 2}}\), but now focus on the statistic \({\rm{R = }}\) the sample range (difference between the largest and smallest values in the sample). Obtain the distribution of\({\rm{R}}\). (Hint: Calculate the value of \({\rm{R}}\) for each outcome and use the probabilities from part (a).)

d. If a random sample of size \({\rm{n = 4}}\) is selected, what is \({\rm{P(\bar X\pounds1}}{\rm{.5)}}\) ? (Hint: You should not have to list all possible outcomes, only those for which\({\rm{\bar x\pounds1}}{\rm{.5}}\).)

Short Answer

Expert verified

(a) The probability distribution of \({\rm{\bar X}}\)

(b) The corresponding probabilities \({\rm{P(\bar X\pounds2}}{\rm{.5) = 0}}{\rm{.85 = 85\% }}\)

(c)

(d) The probability \({\rm{P(\bar x\pounds1}}{\rm{.5) = 0}}{\rm{.24 = 24\% }}\)

Step by step solution

01

Definition

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Obtain the probability distribution of \({\rm{\bar X}}\)

Given:

(a) Determine every random sample containing two data values \({\rm{n = 2}}\) from the set \({\rm{\{ 1,2,3,4\} }}\) (selection of the same value is allowed).

The sample mean is calculated by dividing the total number of values by the number of values.

The product of the probabilities associated with the two data values is its probability.

Make a list of all the sample means from the preceding table.

The sample mean probability is the total of the probabilities in the preceding table that result in the same sample mean.

The probability distribution of \({\rm{\bar X}}\) is shown in this (second) table.

03

Calculating \({\rm{P(\bar X\pounds2}}{\rm{.5)}}\)

(b) Addition rule for disjoint or mutually exclusive events:

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

Add the corresponding probabilities:

\(\begin{aligned}{c}{\rm{P(\bar X\pounds2}}{\rm{.5) = P(X = 1) + P(X = 1}}{\rm{.5) + P(X = 2) + P(X = 2}}{\rm{.5)}}\\{\rm{ = 0}}{\rm{.16 + 0}}{\rm{.24 + 0}}{\rm{.25 + 0}}{\rm{.2}}\\{\rm{ = 0}}{\rm{.85}}\\{\rm{ = 85\% }}\end{aligned}\)

04

Calculating the value of \({\rm{R}}\) for each outcome and use the probabilities

(c) Determine every random sample from the collection \({\rm{1,2,3,4}}\) that has two data values \({\rm{n = 2}}\) (selection of the same value is allowed).

The difference between the biggest and lowest number called the sample range.

The product of the probabilities associated with the two data values is its probability.

Make a list of all the sample ranges from the preceding table.

The sample range probability is the total of the probabilities in the preceding table that result in the same sample mean.

The probability distribution of \({\rm{R}}\) is shown in this (second) table.

05

Calculating the probability

(d) Samples of size \({\rm{n = 4}}\) that will have a sample mean of at most\({\rm{1}}{\rm{.5}}\), need to have the properties that the sum of all data values is at most 6 (because the sample mean is the sum of all data values divided by\({\rm{n = 4}}\)):

\(\begin{aligned}{l}{\rm{\{ (1,1,1,1),(1,1,1,2),(1,1,2,1),(1,2,1,1),(2,1,1,1),(1,1,2,2),(1,2,1,2),}}\\{\rm{(2,1,1,2),(1,2,2,1),(2,1,2,1),(2,2,1,1),(1,1,1,3),(1,1,3,1),(1,3,1,1),(3,1,1,1)\} }}\end{aligned}\)

The probability corresponding to these points is the product of the probability of each of the \({\rm{4}}\) values:

\(\begin{aligned}{l}{\rm{\{ 0}}{\rm{.0256,0}}{\rm{.0192,0}}{\rm{.0192,0}}{\rm{.0192,0}}{\rm{.0192,0}}{\rm{.0144,0}}{\rm{.0144}}\\{\rm{0}}{\rm{.0144,0}}{\rm{.0144,0}}{\rm{.0144,0}}{\rm{.0144,0}}{\rm{.0128,0}}{\rm{.0128,0}}{\rm{.0128,0}}{\rm{.0128\} }}\end{aligned}\)

The probability that \({\rm{\bar x\pounds1}}{\rm{.5}}\) is then the sum of these probabilities:

\(\begin{aligned}{c}{\rm{P(\bar x\pounds1}}{\rm{.5) = 0}}{\rm{.0256 + 4(0}}{\rm{.0192) + 6(0}}{\rm{.0144) + 4(0}}{\rm{.0128)}}\\{\rm{ = 24\% }}\end{aligned}\)

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Most popular questions from this chapter

The difference between the number of customers in line at the express checkout and the number in line at the super-express checkout is\({{\rm{X}}_{\rm{1}}}{\rm{ - }}{{\rm{X}}_{\rm{2}}}\). Calculate the expected difference.

Two different professors have just submitted final exams for duplication. Let \({\rm{X}}\) denote the number of typographical errors on the first professor鈥檚 exam and \({\rm{Y}}\) denote the number of such errors on the second exam. Suppose \({\rm{X}}\) has a Poisson distribution with parameter \({{\rm{\mu }}_{\rm{1}}}\), \({\rm{Y}}\) has a Poisson distribution with parameter \({{\rm{\mu }}_{\rm{2}}}\), and \({\rm{X}}\) and \({\rm{Y}}\) are independent.

a. What is the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\)?

b. What is the probability that at most one error is made on both exams combined?

c. Obtain a general expression for the probability that the total number of errors in the two exams is m (where \({\rm{m}}\) is a nonnegative integer). (Hint: \({\rm{A = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:x + y = m}}} \right\}{\rm{ = }}\left\{ {\left( {{\rm{m,0}}} \right)\left( {{\rm{m - 1,1}}} \right){\rm{,}}.....{\rm{(1,m - 1),(0,m)}}} \right\}\)Now sum the joint pmf over \({\rm{(x,y)}} \in {\rm{A}}\)and use the binomial theorem, which says that

\({\rm{P(X + Y = m)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {\left( {\begin{array}{*{20}{c}}{\rm{m}}\\{\rm{k}}\end{array}} \right){{\rm{a}}^{\rm{k}}}{{\rm{b}}^{{\rm{m - k}}}}{\rm{ = }}\left( {{\rm{a + b}}} \right)} ^{\rm{m}}}\)

In Exercise \({\rm{66}}\), the weight of the beam itself contributes to the bending moment. Assume that the beam is of uniform thickness and density so that the resulting load is uniformly distributed on the beam. If the weight of the beam is random, the resulting load from the weight is also random; denote this load by \({\rm{W}}\) (kip-ft).

a. If the beam is\({\rm{12 }}\)ft long, \({\rm{W}}\)has mean \({\rm{1}}{\rm{.5}}\)and standard deviation \({\rm{.25}}\), and the fixed loads are as described in part (a) of Exercise \({\rm{66}}\), what are the expected value and variance of the bending moment? (Hint: If the load due to the beam were w kip-ft, the contribution to the bending moment would be \(W\grave{o} _0^{12}nxdx\)

b. If all three variables (\({{\rm{X}}_{\rm{1}}}\),\({{\rm{X}}_{\rm{2}}}\), and \({\rm{W}}\)) are normally distributed, what is the probability that the bending moment will be at most \({\rm{200}}\) kip-ft?

You have two lightbulbs for a particular lamp. Let\({\rm{X = }}\)the lifetime of the first bulb and\({\rm{Y = }}\)the lifetime of the second bulb (both in\({\rm{1000}}\)s of hours). Suppose that\({\rm{X}}\)and\({\rm{Y}}\)are independent and that each has an exponential distribution with parameter\({\rm{\lambda = 1}}\).

a. What is the joint pdf of\({\rm{X}}\)and\({\rm{Y}}\)?

b. What is the probability that each bulb lasts at most\({\rm{1000}}\)hours (i.e.,\({\rm{X拢1}}\)and\({\rm{Y拢1)}}\)?

c. What is the probability that the total lifetime of the two bulbs is at most\({\rm{2}}\)? (Hint: Draw a picture of the region before integrating.)

d. What is the probability that the total lifetime is between\({\rm{1}}\)and\({\rm{2}}\)?

The joint pdf of pressures for right and left front tires.

a. Determine the conditional pdf of \({\rm{Y}}\) given that \({\rm{X = x}}\) and the conditional pdf of \({\rm{X}}\) given that \({\rm{Y = y}}\).

b. If the pressure in the right tire is found to be \({\rm{22}}\) psi, what is the probability that the left tire has a pressure of at least \({\rm{25}}\) psi? Compare this to.

c. If the pressure in the right tire is found to be \({\rm{22}}\) psi, what is the expected pressure in the left tire, and what is the standard deviation of pressure in this tire?
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