91Ó°ÊÓ

There have three girls, A, B, and C throw snowballs at a target

Girl A throws 10 times, and the probability that she hits the target on any given throw is 0.3

Girl B throws 15 times, with a 0.2 chance of hitting the goal on each throw.

Girl C throws 20 times, and the probability that she hits the target on any given throw is 0.1

The distribution of the total number of times X that the target is hit will be approximately the normal distribution with the mean,

\(\begin{array}{c}\mu = \left( {10 \times 0.3} \right) + \left( {15 \times 0.2} \right) + \left( {20 \times 0.1} \right)\\ = 3 + 3 + 2\\ = 8\end{array}\)

The variance is,

\(\begin{array}{c}{\sigma ^2} = \left( {10 \times 0.3 \times \left( {1 - 0.3} \right)} \right) + \left( {15 \times 0.2 \times \left( {1 - 0.2} \right)} \right) + \left( {20 \times 0.1 \times \left( {1 - 0.1} \right)} \right)\\ = \left( {10 \times 0.3 \times 0.7} \right) + \left( {15 \times 0.2 \times 0.8} \right) + \left( {20 \times 0.1 \times 0.9} \right)\\ = 2.1 + 2.4 + 1.8\\ = 6.3\end{array}\)

The standard deviation is,

\(\begin{array}{c}\sigma = \sqrt {6.3} \\ = 2.5099\\ \approx 2.51\end{array}\)

Let,

\(\begin{array}{c}Z = \frac{{\left( {X - \mu } \right)}}{\sigma }\\ = \frac{{X - 8}}{{2.51}}\end{array}\)

Z's distribution will resemble a regular normal distribution.

For,\(X = 12\)

\(\begin{array}{c}Z = \frac{{\left( {X - \mu } \right)}}{\sigma }\\ = \frac{{12 - 8}}{{2.51}}\\ = \frac{4}{{2.51}}\\ = 1.5936\end{array}\)

The probability is,

\(\begin{array}{c}P\left( {X \ge 12} \right) = 1 - P\left( {Z < 1.5936} \right)\\ = 1 - 0.9441\\ = 0.0559\end{array}\)

Therefore, the probability that the target will be hit at least 12 times is 0.0559