Respuesta :
Answer:
a) 10.75% probability that exactly 30 of them will pass the quality control checks.
b) Expected value is 32
Variance is 6.4
Step-by-step explanation:
For each monitor, there are only two possible outcomes. Either they will pass the quality checks, or they will not pass these checks. The probability of a monitor passing these checks is independent of other monitors. 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 variance of the binomial distribution is:
[tex]V(X) = np(1-p)[/tex]
80% of the monitors made by your company will pass your stringent quality control checks.
This means that [tex]p = 0.8[/tex]
40 monitors have just come off the assembly line.
This means that [tex]n = 40[/tex]
a) What is the probability that exactly 30 of them will pass the quality control checks?
This is P(X = 30). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 30) = C_{40,30}.(0.8)^{30}.(0.2)^{10} = 0.1075[/tex]
10.75% probability that exactly 30 of them will pass the quality control checks.
b) What are the expected value and variance of the number of these monitors that will pass the quality control checks?
[tex]E(X) = np = 40*0.8 = 32[/tex]
Expected value is 32
[tex]V(X) = np(1-p) = 40*0.8*0.2 = 6.4[/tex]
Variance is 6.4
