Respuesta :
Answer:
10% to decimal is equal to 0.1.
Therefore 20 x 0.1 = 2
Step-by-step explanation:
Each window is independently probable for having minor defects. The expected number of window frames having minor defects = 2.
Other figures are:
- Standard deviation for the number of windows frames to have minor defects in this random sample of 20 window frames = 1.34 approx.
- P(Exactly 2 windows will need minor adjustment) = 0.285 approx.
- P(More than 3 windows will need minor adjustment) = 0.1329 approx.
How to find that a given condition can be modeled by binomial distribution?
Binomial distributions consists of n independent Bernoulli trials.
Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))
Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as
[tex]X \sim B(n,p)[/tex]
The probability that out of n trials, there'd be x successes is given by
[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]
The expected value and variance of X are:
- [tex]E(X) = np[/tex]
- [tex]Var(X) = np(1-p)[/tex]
For the considered case, we can assume that each window is independent of each other for having defects or being perfect.
Each window can be having defects (call it success) or no defects(call it failure).
Then, we can take:
X = number of windows out of 20 windows being perfect (no defect).
Then we get:
- Success = window having defect
- n = 20
- p = 10% = 0.10 (as given in the problem)
- [tex]X \sim B(n=20,p=0.1)[/tex]
Evaluating each sub parts one by one:
- Case a: How many window frames would you expect to have minor defects?
Expect number of defective windows = E(X) = np = 20 × 0.1 = 2
- Case b: Calculate the standard deviation for the number of window frames to have minor defects in this random sample of 20 window frames.
Standard deviation is the positive root of variance.
Standard deviation of number of windows (out of 20 windows of sample) having defects = [tex]\sqrt{Var(X)} = \sqrt{np(1-p)} = \sqrt{2(0.9)} \approx 1.34[/tex]
- Case c: Calculate the probability that exactly 2 window frames will need minor adjustment.
P(X = 2) = [tex]^{20}C_2(0.1)^2(0.9)^{18} \approx 0.285[/tex]
- Case d: What is the probability that more than 3 window frames will need minor adjustments?
P(X > 3) = 1 - P(X ≤ 3)
[tex]P(X > 3) = 1 -[ P(X = 0) + P(X = 1) +P(X= 2)+ P(X = 3)]\\\\P(X > 3) = 1- [^{20}C_0(0.1)^0(0.9)^{20} +^{20}C_1(0.1)^1(0.9)^{19} +\\\: \:\: ^{20}C_2(0.1)^2(0.9)^{18} +^{20}C_3(0.1)^3(0.9)^{17} ]\\\\P(X > 3) \approx 1-[ 0.1216 + 0.2702 + 0.2852 + 0.1901]=1-0.8671=0.1329 \\\\P(X > 3) \approx 0.1329[/tex]
Thus, the needed figures are:
- Expected number of window frames having minor defects = 2
- Standard deviation for the number of windows frames to have minor defects in this random sample of 20 window frames = 1.34 approx.
- P(Exactly 2 windows will need minor adjustment) = 0.285 approx.
- P(More than 3 windows will need minor adjustment) = 0.1329 approx.
Learn more about binomial distribution here:
https://brainly.com/question/13609688
