Question

Using the correction for continuity, determine the probability required in Exercise 6 of Sec. 6.3.

Short answer

Probability that the target will be hit at least 12 times is 0.082

Step-by-step solution

Given information

Girl A throws snowballs at a target 10 times and probability that she will hit the target is 0.3.

Girl B throws snowballs at a target 15 times and probability that she will hit the target is 0.2.

Girl B throws snowballs at a target 20 times and probability that she will hit the target is 0.1.

Calculating the probability of hitting target at least 12 times.

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

\(10\left( {0.3} \right) + 15\left( {0.20} \right) + 20\left( {0.1} \right) = 8\)

Variance is

\(10\left( {0.3} \right)\left( {0.7} \right) + 15\left( {0.2} \right)\left( {0.8} \right) + 20\left( {0.1} \right)\left( {0.8} \right) = 6.3\)

Therefore the distribution of X is approximately a standard normal distribution.

\(\begin{array}{c}Z = \frac{{\left( {X - 8} \right)}}{{\sqrt {6.3} }}\\ = \frac{{\left( {X - 8} \right)}}{{2.51}}\end{array}\)

Thus,

\(\begin{array}{c}{\rm P}\left( {X \ge 12} \right) = {\rm P}\left( {X \ge 11.5} \right)\\ = {\rm P}\left( {Z \ge \frac{{11.5 - 8}}{{2.51}}} \right)\\ = {\rm P}\left( {Z \ge \frac{{3.5}}{{2.51}}} \right)\end{array}\)

\(\begin{array}{l}{\rm P}\left( {X \ge 12} \right) \approx 1 - \phi \left( {1.394} \right)\\{\rm P}\left( {X \ge 12} \right) \approx 0.082\end{array}\)

Therefore,Probability that the target will be hit at least 12 times is 0.082.