Respuesta :
Answer:
a) The approximate probability that more than 25 chips are defective is 0.1075.
b) The approximate probability of having between 20 and 30 defecitve chips is 0.44.
Step-by-step explanation:
Lets call X the total amount of defective chips. X has Binomial distribution with parameters n=1000, p =0.02. Using the Central Limit Theorem, we can compute approximate probabilities for X using a normal variable with equal mean and standard deviation.
The mean of X is np = 1000*0.2 = 20, and the standard deviation is √np(1-p) = √(20*0.98) = 4.427
We will work with a random variable Y with parameters μ=20, σ=4.427. We will take the standarization of Y, W, given by
[tex] W = \frac{Y-\mu}{\sigma} = \frac{Y-20}{4.427} [/tex]
The values of the cummmulative distribution function of the standard normal random variable W, which we will denote [tex] \phi [/tex] , can be found in the attached file. Now we can compute both probabilities. In order to avoid trouble with integer values, we will correct Y from continuity.
a)
[tex]P(X > 25) = P(X > 25.5) \approx P(Y>25.5) = P(\frac{Y-20}{4.427} > \frac{25.5-20}{4.427}) =\\P(W > 1.2423) = 1-\phi(1.2423) = 1-0.8925 = 0.1075[/tex]
Hence the approximate probability that more than 25 chips are defective is 0.1075.
b)
[tex]P(20<X<30) = P(20.5 < X < 29.5) \approx P(20.5<Y>29.5) = \\P(\frac{20.5-20}{4.427} < \frac{Y-20}{4.427} < \frac{29.5-20}{4.427}) = P(0.1129 < W < 2.14) = \phi(2.14)-\phi(0.1129) = \\0.9838-0.5438 = 0.44[/tex]
As a result, the approximate probability of having between 20 and 30 defecitve chips is 0.44.
