Respuesta :
Answer:
0.2345 = 23.45% probability that there are exactly 3 defective resistors in the sample
Step-by-step explanation:
For each resistor, there are only two possible outcomes. Either they are defective, or they are not. The probability of a resistor being defective is independent of other resistors. 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.
8.7% of the resistors produced are defective.
This means that [tex]p = 0.0870[/tex]
If a random sample of 34 resistors is taken, what is the probability that there are exactly 3 defective resistors in the sample?
This is P(X = 3) when n = 34. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{34,3}.(0.087)^{3}.(0.913)^{31} = 0.2345[/tex]
0.2345 = 23.45% probability that there are exactly 3 defective resistors in the sample
