Question

A softball pitcher has a 0.487 probability of throwing a strike for each pitch. If the softball pitcher throws 29 pitches, what is the probability that no more than 14 of them are strikes?

Answer 1

Answer:

0.4801

Step-by-step explanation:

This is a binomial distribution question.

It can be approximated using normal distribution if the following conditions are met:

np > 10

n(1-p) > 10

Here,

n = 29

p = 0.487

So,

np = 14.12

n(1-p) = 14.88

So, we can use normal approximation here:

Binomial:  X ~ B(n,p)  becomes

Normal Approx:  X~ N($np,\sqrt{np(1-p)}$)

Mean is:

$\mu=np=14.123$

Standard Deviation is:

$\sigma=\sqrt{np(1-p)} =2.69$

We need probability of less than or equal to 14, so we can say:

P(x ≤ 14)

Using  $z=\frac{x-\mu}{\sigma}$, we have:

P(x ≤ 14) = $P(\frac{x-\mu}{\sigma} \leq \frac{14-14.123}{2.69})\\=P(z \leq -0.05)\\=0.4801$

Note: We used z table in the last line

So the probability is 0.4801