91Ó°ÊÓ

An average of one-third of the graduating seniors at a particular college have two parents attending the graduation ceremony.

Other third seniors have one parent attend the ceremony.

The remaining one-third of these seniors do not have fathers who attend school.

A specific class has 600 graduating seniors.

For a student chosen at random, the number of parents X who will attend the graduation ceremony has the mean is,

\(\begin{array}{ccccc}\mu = & \frac{0}{3} + \frac{1}{3} + \frac{2}{3}\\ = \frac{{0 + 1 + 2}}{3}\\ = \frac{3}{3}\\ = 1\end{array}\)

The variance is,

\(\begin{array}{c}{\sigma ^2} = E\left[ {{{\left( {X - \mu } \right)}^2}} \right]\\ = \frac{{{{\left( {0 - 1} \right)}^2}}}{3} + \frac{{{{\left( {1 - 1} \right)}^2}}}{3} + \frac{{{{\left( {2 - 1} \right)}^2}}}{3}\\ = \frac{1}{3} + 0 + \frac{1}{3}\\ = \frac{2}{3}\end{array}\)

Therefore, the distribution of the total number of parents W who attend the ceremony will be approximately the normal distribution, with the mean is,

\(\begin{array}{c}\mu = 600 \times 1\\ = 600\end{array}\)

The variance is,

\(\begin{array}{c}{\sigma ^2} = 600 \times \frac{2}{3}\\ = 400\end{array}\)

The standard deviation is,

\(\begin{array}{c}\sigma = \sqrt {400} \\ = 20\end{array}\)

The distribution of Z is,

\(\begin{array}{c}Z = \frac{{W - \mu }}{\sigma }\\ = \frac{{W - 600}}{{20}}\end{array}\)

For

\(W = 650\)

Then,

\(\begin{array}{c}Z = \frac{{W - \mu }}{\sigma }\\ = \frac{{650 - 600}}{{20}}\\ = 2.5\end{array}\)

The probability is,

\(\begin{array}{c}P\left( {W \le 650} \right) = P\left( {Z \le 2.5} \right)\\ = 0.9938\end{array}\)

Therefore, the probability that not more than 650 parents will attend the graduation ceremony is 0.9938