91Ó°ÊÓ

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.

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.