Respuesta :
Answer:
76.10% probability that between 19 and 28 circuits in the sample are defective
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
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 = 90, p = 0.25[/tex]
So
[tex]\mu = E(X) = np = 90*0.25 = 22.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{90*0.25*0.75} = 4.11[/tex]
P(19 ≤ X ≤ 28)
Using continuity correction, this is [tex]P(19-0.5 \leq X \leq 28+0.5) = P(18.5 \leq X \leq 28.5)[/tex], which is the pvalue of Z when X = 28.5 subtracted by the pvalue of Z when X = 18.5. So
X = 28.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{28.5 - 22.5}{4.11}[/tex]
[tex]Z = 1.46[/tex]
[tex]Z = 1.46[/tex] has a pvalue of 0.9279
X = 18.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{18.5 - 22.5}{4.11}[/tex]
[tex]Z = -0.97[/tex]
[tex]Z = -0.97[/tex] has a pvalue of 0.1669
0.9279 - 0.1669 = 0.7610
76.10% probability that between 19 and 28 circuits in the sample are defective
