Respuesta :
Answer:
a) 0.0618 = 6.18% probability that the sample will contain at least three defectives.
b) 0.076 = 7.6% probability that the sample will contain at least three defectives
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
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 distributions 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].
Sample of 20 items is selected from the machine.
This means that [tex]n = 20[/tex]
5% defectives
This means that [tex]p = 0.05[/tex]
(a) the normal approximation to the binomial
The mean is:
[tex]\mu = E(X) = np = 20*0.05 = 1[/tex]
The standard deviation is:
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.05*0.95} = 0.9747[/tex]
The probability is, using continuity correction, [tex]P(X \geq 3 - 2.5) = P(X \geq 2.5)[/tex] , which is 1 subtracted by the pvalue of Z when X = 2.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2.5 - 1}{0.9747}[/tex]
[tex]Z = 1.54[/tex]
[tex]Z = 1.54[/tex] has a pvalue of 0.9382
1 - 0.9382 = 0.0618
0.0618 = 6.18% probability that the sample will contain at least three defectives.
(b) the exact binomial tables
This is:
[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]
In which
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.05)^{0}.(0.95)^{20} = 0.358[/tex]
[tex]P(X = 1) = C_{20,1}.(0.05)^{1}.(0.95)^{19} = 0.377[/tex]
[tex]P(X = 2) = C_{20,2}.(0.05)^{2}.(0.95)^{18} = 0.189[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.358 + 0.377 + 0.189 = 0.924[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.924 = 0.076[/tex]
0.076 = 7.6% probability that the sample will contain at least three defectives
