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Chapter 5: Q74E (page 243)

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 \({\rm{P( - 5拢 X - Y拢 5)}}\) (do not bother with the continuity correction).

Short Answer

Expert verified

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

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:

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

(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

There are two traffic lights on a commuter's route to and from work. Let \({{\rm{X}}_{\rm{1}}}\) be the number of lights at which the commuter must stop on his way to work, and \({{\rm{X}}_{\rm{2}}}\) be the number of lights at which he must stop when returning from work. Suppose these two variables are independent, each with pmf given in the accompanying table (so \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}\) is a random sample of size \({\rm{n = 2}}\)).

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

b. Calculate \({{\rm{\mu }}_{{{\rm{T}}_{\rm{o}}}}}\). How does it relate to \({\rm{\mu }}\), the population mean?

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 \({\rm{V}}\left( {{{\rm{T}}_{\rm{a}}}} \right)\)?

e. Referring back to (d), what are the values of \({\rm{P}}\left( {{{\rm{T}}_{\rm{o}}}{\rm{ = 8}}} \right)\) and \(\text{P}\left( {{\text{T}}_{\text{e}}}\text{ }\!\!{}^\text{3}\!\!\text{ 7} \right)\) (Hint: Don't even think of listing all possible outcomes!)

The lifetime of a certain type of battery is normally distributed with mean value \({\rm{10}}\)hours and standard deviation \({\rm{1}}\)hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only \({\rm{5\% }}\)of all packages?

The manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are\(15,\;30,\;20\)min, respectively, and the standard deviations are\(1,\;2,\;1.5\)min, respectively. What is the probability that it takes at most\(1\)hour of machining time to produce a randomly selected component?

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

There are \({\rm{40}}\) students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of \({\rm{6}}\)min and a standard deviation of \({\rm{6}}\)min.

a. If grading times are independent and the instructor begins grading at \({\rm{6:50}}\) p.m. and grades continuously, what is the (approximate) probability that he is through grading before the \({\rm{11:00}}\) p.m. TV news begins?

b. If the sports report begins at \({\rm{11:10,}}\) what is the probability that he misses part of the report if he waits until grading is done before turning on the TV?
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