In baseball, a . $300$hitter gets a hit in localid="1649737155341" $30\%$of times at bat. When a baseball player hits .localid="1649737159977" $300$, fans tend to be impressed. Typical major leaguers bat about localid="1649737164499" $500$times a season and hit about localid="1649737168930" $0.260$. A hitter's successive tries seem to be independent. Could a typical major leaguer hit .localid="1649737172758" $300$just by chance? Compute an appropriate probability to support your answer.

Number of a baseball hitter$=300$

Percentage of hit times$=30\%$

Typical major leaguers bat number of times$=500$times

There is a binomial distribution difficulty, but we can use a normal distribution to estimate the probability of occurrence.

Since $n=500$and $p=0.260$the binomial distribution has mean

$$\begin{gathered} \mu =np \\ =\left(500(0.260)\right) \\ =130 \end{gathered}$$

and standard deviation

$$\begin{gathered} \sigma =\sqrt{np(1-p)} \\ =\sqrt{500(.260)(.740)} \\ =\sqrt{96.2} \\ \approx 9.8 \end{gathered}$$

To put it another way, major league baseball players average $130$hits per season with a standard deviation of $9.8$hits.

We are interested in whether they can hit localid="1649737229230" $30\%$of the time from 500 at bats, or whether they can get (at least) localid="1649737233775" $500(.300)=150$hits.

We have to find localid="1649737238828" $P(X\ge 150).$

We must utilize the normal distribution to approximate this probability because the conventional binomial distribution is too complex to apply.

It's vital to keep in mind that the normal distribution is a close approximation of the binomial distribution.

localid="1649737245070" $X\ge x$is given by the formula localid="1649737251940" $P(X\ge x)\approx P\left(Z\ge \frac{(x-0.5)-\mu }{\sigma }\right)$so

localid="1650041517648" $$\begin{gathered} P(X\ge 150)\approx P\left(Z\ge \frac{(150-0.5)-130}{\sqrt{96.2}}\right) \\ =P(Z\ge 1.99) \\ =1-P(Z<1.99) \\ =1-.9767 \\ =0.0233 \\ =2.33\mathrm{\%} \end{gathered}$$