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In cost estimation, the total cost of a project is the sum of component task costs. Each of these costs is a random variable with a probability distribution. It is customary to obtain information about the total cost distribution by adding together characteristics of the individual component cost distributions鈥攖his is called the 鈥渞oll-up鈥 procedure. For example, \(E\left( {{X_1} + .. + {X_n}} \right) = E\left( {{X_1}} \right) + .. + E\left( {{X_n}} \right)\), so the roll-up procedure is valid for mean cost. Suppose that there are two component tasks that \({X_1} and {X_2}\)are independent, normally distributed random variables. Is the roll-up procedure valid for the \(7{5^{th}}\) percentile? That is, is the \(7{5^{th}}\) percentile of the distribution of\({X_1} + {X_2}\)the same as the sum of the \(7{5^{th}}\) percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum of percentiles? For what percentiles is the roll-up procedure valid in this case?

Short Answer

Expert verified

When \({\sigma _1} = 0\)or \({\sigma _2} = 0\)or both \({\sigma _1} = {\sigma _2} = 0\) or \({z_\alpha } = 0\). It does not stand for the \({75^{th}}\)percentile for the median.

Step by step solution

01

Definition of Probability Distribution

A probability distribution is a mathematical function in probability theory and statistics that offers the odds of distinct possible outcomes for an experiment.It's a mathematical representation of random phenomena in terms of sample space and event probability.

02

Calculation for the determination of percentile in the given question.

The roll-up procedure is valid when the sum of percentiles is equal to the percentile of sums.

Given random variables are normally distributed! Let the \(\alpha \)percentile of random variable \({X_1}\)be\({\mu _1} + {z_\alpha }{\sigma _1}\) where \({z_\alpha }\;is\;\alpha \)percentile of standard normal distribution, and let the \(\alpha \)percentile of random variable \({X_2}\) be \({\mu _2} + {z_\alpha }{\sigma _2}\)

Because of independence, the \(\alpha \)percentile of a random variable \({X_1} + {X_2}\)is

\(\begin{array}{c}E\left( {{X_1} + {X_2}} \right) + {z_\alpha }\sqrt {V\left( {{X_1} + {X_2}} \right)} = E\left( {{X_1}} \right) + E\left( {{X_2}} \right) + {z_\alpha }\sqrt {V\left( {{X_1}} \right) + V\left( {{X_2}} \right)} \\ = {\mu _1} + {\mu _2} + {z_\alpha }\sqrt {\sigma _1^2 + \sigma _2^2} \end{array}\)

03

Calculation for the determination of percentile in the given question.

Therefore, the sum of percentiles is \({\mu _1} + {z_\alpha }{\sigma _1} + {\mu _2} + {z_\alpha }{\sigma _2}\) and the percentile of the sum is

\({\mu _1} + {\mu _2} + {z_\alpha }\sqrt {\sigma _1^2 + \sigma _2^2} \)

The equality

\({\mu _1} + {z_\alpha }{\sigma _1} + {\mu _2} + {z_\alpha }{\sigma _2} = {\mu _1} + {\mu _2} + {z_\alpha }\sqrt {\sigma _1^2 + \sigma _2^2} \)

stands if \({\sigma _1} = 0\)or both \({\sigma _1} = {\sigma _2} = 0\) or \({z_\alpha } = 0\)

The roll-up procedure does not stand for the \({75^{{\rm{th }}}}\)percentile.

The only percentile for which the roll-up procedure is valid is the median, or \({z_{50}} = 0\).

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Most popular questions from this chapter

A restaurant serves three fixed-price dinners costing \({\rm{12, 15and 20}}\). For a randomly selected couple dining at this restaurant, let \({\rm{X = }}\)the cost of the man鈥檚 dinner and \({\rm{Y = }}\)the cost of the woman鈥檚 dinner. The joint \({\rm{pmf's}}\) of \({\rm{X and Y }}\)is given in the following table:

a. Compute the marginal \({\rm{pmf's}}\)of \({\rm{X and Y }}\)

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c. Are \({\rm{X and Y }}\)independent? Justify your answer.

d. What is the expected total cost of the dinner for the two people?

e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message 鈥淵ou will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen.鈥 How much would the restaurant expect to refund?

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?

A shipping company handles containers in three different sizes:\(\left( 1 \right)\;27f{t^3}\;\left( {3 \times 3 \times 3} \right)\)\(\left( 2 \right) 125 f{t^3}, and \left( 3 \right)\;512 f{t^3}\). Let \({X_i}\left( {i = \;1, 2, 3} \right)\)denote the number of type i containers shipped during a given week. With \({\mu _i} = E\left( {{X_i}} \right)\)and\(\sigma _i^2 = V\left( {{X_i}} \right)\), suppose that the mean values and standard deviations are as follows:

\(\begin{array}{l}{\mu _1} = 200 {\mu _2} = 250 {\mu _3} = 100 \\{\sigma _1} = 10 {\sigma _2} = \,12 {\sigma _3} = 8\end{array}\)

a. Assuming that \({X_1}, {X_2}, {X_3}\)are independent, calculate the expected value and variance of the total volume shipped. (Hint:\(Volume = 27{X_1} + 125{X_2} + 512{X_3}\).)

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\({\rm{Corr(X,Y) = Corr(A,B) \times }}\sqrt {{\rm{Corr}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right)} {\rm{ \times }}\sqrt {{\rm{Corr}}\left( {{{\rm{Y}}_{\rm{1}}}{\rm{,}}{{\rm{Y}}_{\rm{2}}}} \right)} \)

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