91Ó°ÊÓ

The likelihood of an event happening is defined by probability. We may be required to forecast the outcome of an event in a variety of real-life scenarios. The outcomes of an event may or may not be certain. In these instances, we state that the event has a chance of happening or not happening.

Given:

\(\begin{array}{l}\alpha = 50\\\beta = 2\end{array}\)

The mean of a gamma distribution is the product of its variables\(\alpha \;and\;\beta \):

\(\mu = E(X) = \alpha \beta = 50(2) = 100\)

The variance of a gamma distribution is the product of its variables squared\(\alpha \;and\;\beta \):

\({\sigma ^2} = V(X) = \alpha {\beta ^2} = 50{(2)^2} = 50(4) = 200\)

The standard deviation is the square root of the variance:

\(\sigma = \sqrt {200} = 10\sqrt 2 \approx 14.1421\)

Given is that it is appropriate to approximate the gamma distribution with the normal distribution. We will approximate the gamma distribution with the normal distribution of mean \(\mu = 100\)and standard deviation

\(\sigma = 10\sqrt 2 \approx 14.1421\)

Note: No continuity correction is required, since gamma distribution is a continuous function and the normal distribution is also a continuous function.

The z-score is the value decreased by the mean and divided by the standard deviation.

\(z = \frac{{x - np}}{{\sqrt {np(1 - p)} }} = \frac{{125 - 100}}{{10\sqrt 2 }} \approx 1.77\)

Determine the corresponding normal probability using table.

\(P(X \le 125) = P(Z < 1.77) = 0.9616 = 96.16\% \)