Respuesta :
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:
[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 = 85, p = 0.21[/tex].
So
[tex]E(X) = np = 85*0.21 = 17.85[/tex]
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{85*0.21*0.79} = 3.7552[/tex]
P ( 14 ≤ X ≤ 20 )
Using continuity correction, this is [tex]P(14 - 0.5 \leq X \leq 20 + 0.5) = P(13.5 \leq X \leq 20.5)[/tex], which is the pvalue of Z when X = 20.5 subtracted by the pvalue of Z when X = 13.5. So
X = 20.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.5 - 17.85}{3.7552}[/tex]
[tex]Z = 0.71[/tex]
[tex]Z = 0.71[/tex] has a pvalue of 0.7611
X = 13.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{13.5 - 17.85}{3.7552}[/tex]
[tex]Z = -1.16[/tex]
[tex]Z = -1.16[/tex] 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.
