Respuesta :
Answer:
41.68% probability that a randomly selected batch contains more than 6 defects.
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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 120, p = 0.05[/tex]
So
[tex]\mu = E(X) = np = 120*0.05 = 6[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.05*0.95} = 2.39[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{6.5 - 6}{2.39}[/tex]
[tex]Z = 0.21[/tex]
[tex]Z = 0.21[/tex] has a pvalue of 0.5832
1 - 0.5832 = 0.4168
41.68% probability that a randomly selected batch contains more than 6 defects.
