Answer:
63.81% probability that between 14 and 20 circuits in the sample are defective.
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:
.
So
![E(X) = np = 85*0.21 = 17.85](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np%20%3D%2085%2A0.21%20%3D%2017.85)
![\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{85*0.21*0.79} = 3.7552](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B85%2A0.21%2A0.79%7D%20%3D%203.7552)
P ( 14 ≤ X ≤ 20 )
Using continuity correction, this is
, which is the pvalue of Z when X = 20.5 subtracted by the pvalue of Z when X = 13.5. So
X = 20.5
![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{20.5 - 17.85}{3.7552}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B20.5%20-%2017.85%7D%7B3.7552%7D)
![Z = 0.71](https://tex.z-dn.net/?f=Z%20%3D%200.71)
has a pvalue of 0.7611
X = 13.5
![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{13.5 - 17.85}{3.7552}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B13.5%20-%2017.85%7D%7B3.7552%7D)
![Z = -1.16](https://tex.z-dn.net/?f=Z%20%3D%20-1.16)
has a pvalue of 0.1230
0.7611 - 0.1230 = 0.6381
63.81% probability that between 14 and 20 circuits in the sample are defective.