Question

P(X< ) 1-P(X> ) A softball pitcher has a 0.626 probability of throwing a strike for each curve ball pitch. If the softball pitcher throws 30 curve balls, what is the probability that no more than 16 of them are strikes? Fill in the blanks below to represent the probability

Answer 1

Answer:

19.49% probability that no more than 16 of them are strikes

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 30, p = 0.626$

So

$\mu = E(X) = np = 30*0.626 = 18.78$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65$

What is the probability that no more than 16 of them are strikes?

Using continuity correction, this is $P(X \leq 16 + 0.5) = P(X \leq 16.5)$, which is the pvalue of Z when X = 16.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16.5 - 18.78}{2.65}$

$Z = -0.86$

$Z = -0.86$ has a pvalue of 0.1949

19.49% probability that no more than 16 of them are strikes