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

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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

Mathematics

1 answer:

Inessa [12.5K] · Answer 1 · 2026-04-03T08:16:55+03:00

Answer:

The likelihood is 19.49% that no more than 16 of them are strikes

Step-by-step explanation:

Using the binomial probability distribution

The probability of a specific number of successes in n repeated trials, with a given probability p.

This can be approximated to a normal distribution, utilizing the expected value and standard deviation.

The expected value for the binomial distribution is:

$E(X) = np$

The standard deviation for the binomial distribution is:

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

For normal probability distributions

Problems related to normally distributed samples can be analyzed utilizing the z-score formula.

In a dataset characterized by a mean $\mu$ and a standard deviation $\sigma$ , the z-score for a measure X can be calculated as:

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

The z-score indicates how many standard deviations a measure is from the average. After computing the z-score, one can reference the z-score table to find the associated p-value. This p-value represents the probability that the measure's value is less than X, thus indicating the percentile of X. By subtracting this p-value from 1, we derive the likelihood that the measure's value surpasses X.

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