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91Ó°ÊÓ

In an area having sandy soil,\({\rm{50}}\)small trees of a certain type were planted, and another \({\rm{50}}\) trees were planted in an area having clay soil. Let \({\rm{X = }}\) the number of trees planted in sandy soil that survive \({\rm{1}}\) year and \({\rm{Y = }}\)the number of trees planted in clay soil that survive \({\rm{1}}\) year. If the probability that a tree planted in sandy soil will survive \({\rm{1}}\)year is \({\rm{.7}}\)and the probability of \({\rm{1}}\)-year survival in clay soil is \({\rm{.6}}\), compute an approximation to \({\rm{P( - 5£ X - Y£ 5)}}\) (do not bother with the continuity correction).

Short Answer

Expert verified

\({\rm{P( - 5£ X - Y£ 5) = 0}}{\rm{.4826}}\)

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

Determining the approximation to \({\rm{P( - 5£ X - Y£ 5)}}\)

Because an approximation of the probability is required, notice that random variables \({\rm{X}}\)and \({\rm{Y}}\)have a normal distribution. The mean value of the random variable \({\rm{X}}\)is

\({{\rm{\mu }}_{\rm{X}}}{\rm{ = 50 \times 0}}{\rm{.7 = 35}}\)

where \({\rm{50 }}\)is the number of little trees and the likelihood of a tree planted in sandy soil surviving a year is \({\rm{0}}{\rm{.7}}\), it is estimated that\({\rm{35}}\)total trees will survive (the expected value for a Binomial distribution is \({\rm{n \times p}}\)). The random variable \({\rm{Y}}\)has the same mean value.

\({{\rm{\mu }}_{\rm{Y}}}{\rm{ = 50 \times 0}}{\rm{.6 = 30}}\)

where \({\rm{50 }}\) is the number of little trees and the probability of a tree planted in clay soil surviving a year is \({\rm{0}}{\rm{.6 }}\), a total of\({\rm{30}}\)trees should survive.

The random variable \({\rm{X}}\)has a variance of

\({\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = 50 \times 0}}{\rm{.7 \times (1 - 0}}{\rm{.7) = 10}}{\rm{.5}}\)

\({\rm{n \times p \times (1 - p)}}\)is used to calculate the variance (as for Binomial distribution). The variance of the random variable \({\rm{Y}}\)is similar.

\({\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ = 50 \times 0}}{\rm{.6 \times (1 - 0}}{\rm{.6) = 12}}\)

The random variable \({\rm{X - Y}}\)has an expected value of

\({{\rm{\mu }}_{{\rm{X - Y}}}}{\rm{ = E(X - Y) = E(X) - E(Y) = 35 - 30 = 5}}\)

The random variable \({\rm{X - Y}}\)has a variance of

\({\rm{\sigma }}_{{\rm{X - Y}}}^{\rm{2}}{\rm{ = }}{{\rm{1}}^{\rm{2}}}{\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ + ( - 1}}{{\rm{)}}^{\rm{2}}}{\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ = 10}}{\rm{.5 + 12 = 22}}{\rm{.5}}\)

Finally, using a standard normal distribution, the requested probability can be estimated as follows:

\(\begin{array}{*{20}{c}}{{\rm{P( - 5£ X - Y£ 5)}}}&{{\rm{ = P}}\left( {\frac{{{\rm{ - 5 - 5}}}}{{\sqrt {{\rm{22}}{\rm{.5}}} }}{\rm{£ }}\frac{{{\rm{X - Y - }}{{\rm{\mu }}_{{\rm{X - Y}}}}}}{{{{\rm{\sigma }}_{{\rm{X - Y}}}}}}{\rm{£ }}\frac{{{\rm{5 - 5}}}}{{\sqrt {{\rm{22}}{\rm{.5}}} }}} \right)}\\{}&{{\rm{ = P( - 2}}{\rm{.11£ Z£ 0) = P(Z£ 0) - P(Z£ - 2}}{\rm{.11)}}}\\{}&{\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0}}{\rm{.5 - 0}}{\rm{.0174}}}\\{}&{{\rm{ = 0}}{\rm{.4826}}}\end{array}\)

(1): from the appendix's normal probability table. Software can also be used to calculate the likelihood.

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Most popular questions from this chapter

Refer to Exercise \({\rm{46}}\). Suppose the distribution is normal (the cited article makes that assumption and even includes the corresponding normal density curve).

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a. Determine the pmf of \({{\rm{T}}_{\rm{o}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}\).

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c. Calculate \({\rm{\sigma }}_{{{\rm{T}}_{\rm{o}}}}^{\rm{2}}\). How does it relate to \({{\rm{\sigma }}^{\rm{2}}}\), the population variance?

d. Let \({{\rm{X}}_{\rm{3}}}\) and \({{\rm{X}}_{\rm{4}}}\) be the number of lights at which a stop is required when driving to and from work on a second day assumed independent of the first day. With \({{\rm{T}}_{\rm{o}}}{\rm{ = }}\) the sum of all four \({{\rm{X}}_{\rm{i}}}\) 's, what now are the values of \({\rm{E}}\left( {{{\rm{T}}_{\rm{a}}}} \right)\) and \({\rm{V}}\left( {{{\rm{T}}_{\rm{a}}}} \right)\)?

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a. Let \({{\rm{X}}_{\rm{1}}}\)have a chi-squared distribution with parameter \({{\rm{\nu }}_{\rm{1}}}\) (see Section 4.4), and let \({{\rm{X}}_{\rm{2}}}\)be independent of \({{\rm{X}}_{\rm{1}}}\)and have a chi-squared distribution with parameter\({{\rm{v}}_{\rm{2}}}\). Use the technique of to show that \({{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}\)has a chi-squared distribution with parameter\({{\rm{v}}_{\rm{1}}}{\rm{ + }}{{\rm{v}}_{\rm{2}}}\).

b. You were asked to show that if \({\rm{Z}}\)is a standard normal \({\rm{rv}}\), then \({{\rm{Z}}^{\rm{2}}}\)has a chi squared distribution with\({\rm{v = 1}}\). Let \({{\rm{Z}}_{\rm{1}}}{\rm{,}}{{\rm{Z}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{Z}}_{\rm{n}}}\)be \({\rm{n}}\)independent standard normal \({\rm{rv}}\) 's. What is the distribution of\({\rm{Z}}_{\rm{1}}^{\rm{2}}{\rm{ + }}...{\rm{ + Z}}_{\rm{n}}^{\rm{2}}\)? Justify your answer.

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

where \({{\rm{X}}_{\rm{1}}}\)and \({{\rm{X}}_{\rm{2}}}\)are replicate measurements on the value of\({\rm{A}}\), and \({{\rm{Y}}_{\rm{1}}}\)and \({{\rm{Y}}_{\rm{2}}}\)are defined analogously with respect to\({\rm{B}}\). What effect does the presence of measurement error have on the correlation?

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