Respuesta :
Answer:
0.44% approximate probability that in this week more than 80 panels have dead pixels.
Step-by-step explanation:
For each LCD panel, there are only two possible outcomes. Either it has a dead pixel, or it does not. So we use the binomial probability distribution to solve this problem.
However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal, by the Central Limit Theorem(CLT).
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]p = 0.03, n = 2000[/tex]
So
[tex]\mu = np = 2000*0.03 = 60[/tex]
[tex]\sigma = \sqrt{np(1-p)} = \sqrt{2000*0.03*0.97} = 7.63[/tex]
Using the CLT, what is the approximate probability that in this week more than 80 panels have dead pixels?
This is 1 subtracted by the pvalue of Z when X = 80. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 60}{7.63}[/tex]
[tex]Z = 2.62[/tex]
[tex]Z = 2.62[/tex] has a pvalue of 0.9956.
So there is a 1-0.9956 = 0.0044 = 0.44% approximate probability that in this week more than 80 panels have dead pixels.
