Respuesta :
Answer:
a) The mean is 3 computers and the standard deviation is 1.73 computers.
b) 0.916
c) 0.04971
Step-by-step explanation:
For each computer, there are only two possible outcomes. Either they experience failure during the warranty period, or they do not. The probability of a computer experiencing a failure is independent of other computers. So we use the binomial probability distribution to solve this question.
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]
In this problem we have that:
[tex]n = 3000, p = 0.001[/tex]
(a) What are the expected value and standard deviation of the number of computers in the sample that have the defect? (Round your standard deviation to two decimal places.)
[tex]E(X) = np = 3000*0.001 = 3[/tex]
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{3000*0.001*0.999} = 1.73[/tex]
The mean is 3 computers and the standard deviation is 1.73 computers.
(b) What is the (approximate) probability that more than 5 sampled computers have the defect? (Round your answer to three decimal places.)
Either 5 or less have the defect, or more than five do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 5) + P(X > 5) = 1[/tex]
We want P(X > 5).
So
[tex]P(X > 5) = 1 - P(X \leq 5)[/tex]
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3000,0}.(0.001)^{0}.(0.999)^{3000} = 0.05[/tex]
[tex]P(X = 1) = C_{3000,1}.(0.001)^{1}.(0.999)^{2999} = 0.149[/tex]
[tex]P(X = 2) = C_{3000,2}.(0.001)^{2}.(0.999)^{2998} = 0.224[/tex]
[tex]P(X = 3) = C_{3000,3}.(0.001)^{3}.(0.999)^{2997} = 0.224[/tex]
[tex]P(X = 4) = C_{3000,4}.(0.001)^{4}.(0.999)^{2996} = 0.168[/tex]
[tex]P(X = 5) = C_{3000,5}.(0.001)^{5}.(0.999)^{2995} = 0.101[/tex]
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.05 + 0.149 + 0.224 + 0.224 + 0.168 + 0.101 = 0.916[/tex]
(c) What is the (approximate) probability that no sampled computers have the defect? (Round your answer to five decimal places.)
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3000,0}.(0.001)^{0}.(0.999)^{3000} = 0.04971[/tex]
