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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

Question: The number of customers waiting for gift-wrap service at a department store is an rv X with possible values \({\rm{0,1,2,3,4}}\)and corresponding probabilities \({\rm{.1,}}{\rm{.2,}}{\rm{.3,}}{\rm{.25,}}{\rm{.15}}{\rm{.}}\)A randomly selected customer will have \({\rm{1,2}}\),or \({\rm{3}}\) packages for wrapping with probabilities \({\rm{.6,}}{\rm{.3,}}\)and \({\rm{.1,}}\)respectively. Let \({\rm{Y = }}\)the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer).

a. Determine \({\rm{P(X = 3,Y = 3)}}\), i.e., \({\rm{P(3,3)}}\).

b. Determine \({\rm{p(4,11)}}\).

Suppose the amount of liquid dispensed by a certain machine is uniformly distributed with lower limit \({\rm{A = 8oz}}\) and upper limit\({\rm{B = 10oz}}\). Describe how you would carry out simulation experiments to compare the sampling distribution of the (sample) fourth spread for sample sizes\({\rm{n = 5,10,20}}\), and\({\rm{30}}\).

The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value \({\rm{10}}\) min and standard deviation \({\rm{2}}\) min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most \({\rm{11}}\) min?

Consider a system consisting of three components as pictured. The system will continue to function as long as the first component functions and either component \({\rm{2}}\) or component \({\rm{3}}\)functions. Let \({{\rm{X}}_{{\rm{1,}}}}{{\rm{X}}_{\rm{2}}}\), and \({{\rm{X}}_{\rm{3}}}\) denote the lifetimes of components \({\rm{1}}\), \({\rm{2}}\), and \({\rm{3}}\), respectively. Suppose the \({{\rm{X}}_{\rm{i}}}\) 鈥檚 are independent of one another and each \({{\rm{X}}_{\rm{i}}}\) has an exponential distribution with parameter \({\rm{\lambda }}\).

a. Let \({\rm{Y}}\) denote the system lifetime. Obtain the cumulative distribution function of \({\rm{Y}}\)and differentiate to obtain the pdf. (Hint: \({{\rm{F}}_{\left( {\rm{Y}} \right)}}{\rm{P}}\left\{ {{\rm{Y}} \le {\rm{y}}} \right\}\); express the event \(\left\{ {{\rm{Y}} \le {\rm{y}}} \right\}\)in terms of unions and/or intersections of the three events \(\left\{ {{{\rm{X}}_{\rm{i}}} \le {\rm{y}}} \right\}\), \(\left\{ {{{\rm{X}}_{\rm{2}}} \le {\rm{y}}} \right\}\), and \(\left\{ {{{\rm{X}}_3} \le {\rm{y}}} \right\}\).)

b. Compute the expected system lifetime

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.
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