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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

Suppose the expected tensile strength of type-A steel is \({\rm{105ksi}}\)and the standard deviation of tensile strength is \({\rm{8ksi}}\). For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are \({\rm{100ksi}}\)and \({\rm{6ksi}}\), respectively. Let \({\rm{\bar X = }}\)the sample average tensile strength of a random sample of \({\rm{40}}\) type-A specimens, and let \({\rm{\bar Y = }}\)the sample average tensile strength of a random sample of \({\rm{35}}\)type-B specimens.

a. What is the approximate distribution of \({\rm{\bar X ? of \bar Y?}}\)

b. What is the approximate distribution of \({\rm{\bar X - \bar Y}}\)? Justify your answer.

c. Calculate (approximately) \(P( - 1拢\bar X - \bar Y拢1)\)

d. Calculate. If you actually observed , would you doubt that \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}{\rm{ = 5?}}\)

A stockroom currently has \({\rm{30}}\) components of a certain type, of which \({\rm{8}}\) were provided by supplier \({\rm{1,10}}\)by supplier \({\rm{2}}\) , and \({\rm{12}}\) by supplier \({\rm{3}}\). Six of these are to be randomly selected for a particular assembly. Let \({\rm{X = }}\) the number of supplier l's components selected, \({\rm{Y = }}\) the number of supplier \({\rm{2}}\) 's components selected, and \({\rm{p(x,y)}}\) denote the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\).

a. What is \({\rm{p(3,2)}}\) ? (Hint: Each sample of size \({\rm{6}}\) is equally likely to be selected. Therefore, \({\rm{p(3,2) = }}\) (number of outcomes with \({\rm{X = 3}}\) and \({\rm{Y = 2)/(}}\) total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.)

b. Using the logic of part (a), obtain \({\rm{p(x,y}}\) ). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories.)

Answer

Garbage trucks entering a particular waste-management facility are weighed prior to offloading their contents. Let \({\rm{X = }}\)the total processing time for a randomly selected truck at this facility (waiting, weighing, and offloading). The article "Estimating Waste Transfer Station Delays Using GPS" (Waste Mgmt., \({\rm{2008: 1742 - 1750}}\)) suggests the plausibility of a normal distribution with mean \({\rm{13\;min}}\)and standard deviation \({\rm{4\;min}}\)for\({\rm{X}}\). Assume that this is in fact the correct distribution.

a. What is the probability that a single truck's processing time is between \({\rm{12}}\) and \({\rm{15\;min}}\)?

b. Consider a random sample of \({\rm{16}}\) trucks. What is the probability that the sample mean processing time is between \({\rm{12}}\) and\({\rm{15\;min}}\)?

c. Why is the probability in (b) much larger than the probability in (a)?

d. What is the probability that the sample mean processing time for a random sample of \({\rm{16}}\) trucks will be at least\({\rm{20\;min}}\)?

Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv鈥檚 with expected values \({\mu _1}, {\mu _2}, and {\mu _3}\)and variances \(\sigma _1^2 , \sigma _2^2, and \sigma _3^2 \), respectively. a. If \(\mu = {\mu _2} = {\mu _3} = 60\)and\(\sigma _1^2 = \sigma _2^2 = \sigma _3^2 = 15\), calculate \(P\left( {{T_0} \le 200} \right)\)and\(P\left( {150 \le {T_0} \le 200} \right)\)? b. Using the \(\mu 's and \sigma 's\)given in part (a), calculate both \(P\left( {55 \le X} \right)\)and \(P\left( {58 \le X \le 62} \right)\).c. Using the \(\mu 's and \sigma 's\)given in part (a), calculate and interpret\(P\left( { - 10 \le {X_1} - .5{X_2} - .5{X_3} \le 5} \right)\). d. If\({\mu _1} = 40, {\mu _1} = 50, {\mu _1} = 60,\),\( \sigma _1^2 = 10, \sigma _2^2 = 12, and \sigma _3^2 = 14\) calculate \(P\left( {{X_1} + {X_2} + {X_3} \le 160} \right)\)and also \(P\left( {{X_1} + {X_2} \ge 2{X_3}} \right).\)

A binary communication channel transmits a sequence of 鈥渂its鈥� (\({\bf{0s}}{\rm{ }}{\bf{and}}{\rm{ }}{\bf{1s}}\)). Suppose that for any particular bit transmitted, there is a\({\bf{10}}\% \)chance of a transmission error (a zero becoming a one or a one becoming a zero). Assume that bit errors occur independently of one another.

a. Consider transmitting\(1000\)bits. What is the approximate probability that at most\(125\)transmission errors occur?

b. Suppose the same\(1000\)-bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within\(50\)of the number of errors in the second?
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