Automobiles Fail Emissions Inspection

Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection.

a. Among 15 randomly selected cars, what is the probability that at most 5 fail the inspection?

b. Among 15 randomly selected cars, what is the probability that between 5 and 10 (inclusive) fail to pass inspection?

c. Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation of the number that pass inspection?

d. What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?

a. The binomial probability can be calculated using the formula P(x) = nCx * p^x * (1-p)^n-x. Here, n=15, x=5, and p=0.3. Thus,

P(X ≤ 5) = 15C5 * 0.3^5 * (1-0.3)^10 = 0.719.

Therefore, the probability that at most 5 fail the inspection is 0.719.

b. The probability that between 5 and 10 (inclusive) fail to pass inspection is 0.462.

c. Let Y be the number of cars that pass the inspection among 25 randomly selected cars. Then, Y follows a binomial distribution with n=25 and p=0.7 (since 70% pass the inspection). The mean value of Y is E(Y) = np = 25 x 0.7 = 17.5. The standard deviation of Y is σ = sqrt(np(1-p)) = sqrt(25 x 0.7 x 0.3) = approx. 2.291.

d. To find the probability that Y is within 1 standard deviation of the mean value, we need to calculate P(E(Y)-σ ≤ Y ≤ E(Y)+σ). Using a normal distribution approximation and the values from part c, we can standardize Y as follows: Z = (Y - E(Y))/σ Then, we need to find P(-1 ≤ Z ≤ 1). Using a standard normal table or a calculator, we get P(-1 ≤ Z ≤ 1) = 0.6827. Therefore, the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value is 0.6827.