Answer:
0.9726 = 97.26% approximate probability that X is at most 30
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 distributions 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
,
.
11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped).
This means that ![p = 0.11](https://tex.z-dn.net/?f=p%20%3D%200.11)
Random sample of 200 shafts
This means that ![n = 200](https://tex.z-dn.net/?f=n%20%3D%20200)
Mean and Standard deviation:
![\mu = E(x) = np = 200*0.11 = 22](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28x%29%20%3D%20np%20%3D%20200%2A0.11%20%3D%2022)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.11*0.89} = 4.42](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%7B200%2A0.11%2A0.89%7D%20%3D%204.42)
(a) What is the (approximate) probability that X is at most 30
Using continuity correction, this is
, which is the pvalue of Z when X = 30.5. 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{30.5 - 22}{4.42}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B30.5%20-%2022%7D%7B4.42%7D)
![Z = 1.92](https://tex.z-dn.net/?f=Z%20%3D%201.92)
has a pvalue of 0.9726.
0.9726 = 97.26% approximate probability that X is at most 30