91影视

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

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

and standard deviation

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

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" $P(X鈮�150).$

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" $X鈮�x$is given by the formula localid="1649737251940" $P(X鈮�x)鈮�P\left(Z鈮�\frac{(x-0.5)-渭}{蟽}\right)$so

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