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](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
,
.
In this problem, we have that:
![n = 30, p = 0.626](https://tex.z-dn.net/?f=n%20%3D%2030%2C%20p%20%3D%200.626)
So
![\mu = E(X) = np = 30*0.626 = 18.78](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%2030%2A0.626%20%3D%2018.78)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{30*0.626*0.374} = 2.65](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B30%2A0.626%2A0.374%7D%20%3D%202.65)
What is the probability that no more than 16 of them are strikes?
Using continuity correction, this is
, which is the pvalue of Z when X = 16.5. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{16.5 - 18.78}{2.65}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B16.5%20-%2018.78%7D%7B2.65%7D)
![Z = -0.86](https://tex.z-dn.net/?f=Z%20%3D%20-0.86)
has a pvalue of 0.1949
19.49% probability that no more than 16 of them are strikes