Respuesta :
Using the binomial distribution, it is found that:
a) 0.1271 = 12.71% probability of obtaining at least three defects in the sample, if the entire shipment has 5% defectives. 5% defective is between 1 and 2 defects, but since this probability is more than the unlikely threshold 5%, more than 2 defects could still mean that Intel has provided a shipment which is consistent with the contract.
b) The mean is of 1.25 defective chips and the standard deviation is of 1.09 defective chips.
c) 0.8908 = 89.08% probability that more than 17 chips in the sample are good.
For each chip, there are only two possible outcomes, either it is defective, or it is not. The probability of a chip being defective is independent of any other chip, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 25 chips are tested, thus, [tex]n = 25[/tex].
Item a:
Supposing 5% defective, we have that [tex]p = 0.05[/tex].
The probability 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]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{25,0}.(0.05)^{0}.(0.95)^{25} = 0.2774[/tex]
[tex]P(X = 1) = C_{25,1}.(0.05)^{1}.(0.95)^{24} = 0.3650[/tex]
[tex]P(X = 2) = C_{25,2}.(0.05)^{2}.(0.95)^{23} = 0.2305[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2774 + 0.3650 + 0.2305 = 0.8729[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.8729 = 0.1271[/tex]
0.1271 = 12.71% probability of obtaining at least three defects in the sample, if the entire shipment has 5% defectives. 5% defective is between 1 and 2 defects, but since this probability is more than the unlikely threshold 5%, more than 2 defects could still mean that Intel has provided a shipment which is consistent with the contract.
Item b:
The mean is:
[tex]E(X) = np = 25(0.05) = 1.25[/tex]
The standard deviation is:
[tex]\sqrt{V(X)} = \sqrt{np(1 - p)} = \sqrt{25(0.05)(0.95)} = 1.09[/tex]
The mean is of 1.25 defective chips and the standard deviation is of 1.09 defective chips.
Item c:
20% are defective, so 80% are good, which means that [tex]p = 0.8[/tex]
The probability is:
[tex]P(X > 17) = P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25)[/tex]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 18) = C_{25,18}.(0.8)^{18}.(0.2)^{7} = 0.1108[/tex]
[tex]P(X = 19) = C_{25,19}.(0.8)^{19}.(0.2)^{6} = 0.1633[/tex]
[tex]P(X = 20) = C_{25,20}.(0.8)^{20}.(0.2)^{5} = 0.1960[/tex]
[tex]P(X = 21) = C_{25,21}.(0.8)^{21}.(0.2)^{4} = 0.1867[/tex]
[tex]P(X = 22) = C_{25,22}.(0.8)^{22}.(0.2)^{3} = 0.1358[/tex]
[tex]P(X = 23) = C_{25,23}.(0.8)^{23}.(0.2)^{2} = 0.0708[/tex]
[tex]P(X = 24) = C_{25,24}.(0.8)^{24}.(0.2)^{1} = 0.0236[/tex]
[tex]P(X = 25) = C_{25,25}.(0.8)^{25}.(0.2)^{0} = 0.0038[/tex]
Then:
[tex]P(X > 17) = P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) + P(X = 24) + P(X = 25) = 0.1108 + 0.1633 + 0.1960 + 0.1867 + 0.1358 + 0.0708 + 0.0236 + 0.0038 = 0.8908[/tex]
0.8908 = 89.08% probability that more than 17 chips in the sample are good.
A similar problem is given at https://brainly.com/question/24863377
