Question

You may need to use the appropriate appendix tabie or tectunology to aniwer this question. in a seuthern take, it was revealed that 9% of all autemobiles in the state did not pass inspection. (a) of the next ten automobiles entering the inspection statien, what is the probabily that none will pass inspection? (b) of the next ton automobiles entering the inspection station, what is the probabily that all will pass inspection? (c) of the nent ten automobiies entenng the inspection station, what is the probability that exactyy two will not pass inspection? (d) Of the next tan autemobiles entering the inspection station, what is the grobecaity that move than three eili not pass inspection? (e) of the next ten automobiles entering the insooction itatien, what is the probabiity that fewer than two will not pass inspection? (f) Or the next ten automebies entering the inspection station, find the expected number of automobiles not pasing inspection. (9) of the next ten autemoblies entering the inspection station, determine the stansiand devation for the number of cars not passing inseeces

Answer 1

(a) The probability that none of the next ten automobiles will pass inspection is approximately 0.0000003874, or 0.00003874%.

(b) The probability that all of the next ten automobiles will pass inspection is approximately 0.3487, or 34.87%.

Step-by-step explanation:

(a) To find the probability that none of the next ten automobiles will pass inspection, we use the binomial probability formula: P(X = k) = C(n, k) * p^k * (1 - p)^(n - k), where n is the number of trials, k is the number of successes, p is the probability of success, and C(n, k) is the binomial coefficient. For this case, n = 10, k = 0, and p = 0.09. Plugging in these values, we get P(X = 0) ≈ 0.0000003874.

(b) Similarly, to find the probability that all of the next ten automobiles will pass inspection, we use the same formula with k = 10. P(X = 10) ≈ 0.3487.

(c) To calculate the probability that exactly two out of the next ten automobiles will not pass inspection, we use the binomial probability formula with k = 2. P(X = 2) ≈ 0.0024.

(d) For the probability that more than three out of the next ten automobiles will not pass inspection, we find P(X > 3) by summing the probabilities for k = 4, 5, ..., 10. The result is approximately 0.2015.

(e) The probability that fewer than two out of the next ten automobiles will not pass inspection is P(X < 2), which is equal to P(X = 0) + P(X = 1). The sum is approximately 0.0000007453.

(f) To find the expected number of automobiles not passing inspection, we use the formula E(X) = np, where n is the number of trials and p is the probability of success. Here, E(X) ≈ 0.9.

(g) To determine the standard deviation for the number of cars not passing inspection, we use the formula σ = sqrt(np(1-p)), where σ is the standard deviation. Plugging in the values, we get σ ≈ 0.3.