Answer:
41.68% probability that a randomly selected batch contains more than 6 defects.
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://img.qammunity.org/2021/formulas/mathematics/college/66n16kmn896qth698tyf6rfu48vhaipkmv.png)
The standard deviation of the binomial distribution is:
![√(V(X)) = √(np(1-p))](https://img.qammunity.org/2021/formulas/mathematics/college/50rvo6hmelacol69fy9pzbmom4zmpsvsnd.png)
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 = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
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 = 120, p = 0.05](https://img.qammunity.org/2021/formulas/mathematics/college/76ncqm6cfesse8gixsnnn6u8socnjnr4jl.png)
So
![\mu = E(X) = np = 120*0.05 = 6](https://img.qammunity.org/2021/formulas/mathematics/college/6ebblrj5m95o0v6m9e8p8yq72u5jordfan.png)
![\sigma = √(V(X)) = √(np(1-p)) = √(120*0.05*0.95) = 2.39](https://img.qammunity.org/2021/formulas/mathematics/college/l899iobzbcrygxxoj2k8dy24et1ikij27k.png)
Estimate the probability that a randomly selected batch contains more than 6 defects.
Using continuity correction, this is P(X > 6 + 0.5) = P(X > 6.5), which is 1 subtracted by the pvalue of Z when X = 6.5. So
![Z = (X - \mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/college/c62rrp8olhnzeelpux1qvr89ehugd6fm1f.png)
![Z = (6.5 - 6)/(2.39)](https://img.qammunity.org/2021/formulas/mathematics/college/sp0q7lv4pzayzllohhhdet8xj4lo2az675.png)
![Z = 0.21](https://img.qammunity.org/2021/formulas/mathematics/college/syuvp2t5sxkvedyw2rtrucw0kiyrxjg6tz.png)
has a pvalue of 0.5832
1 - 0.5832 = 0.4168
41.68% probability that a randomly selected batch contains more than 6 defects.