Question

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?

Short answer

a) Probability for at most 5 cars failing is calculated using binomial distribution. b) For 5 to 10 cars failing, use the same concept as a) and subtract the probabilities. c) Use the formula for mean and standard deviation for binomial distribution. d) Convert into standard normal variable and use standard distribution table.

Step-by-step solution

Calculate probability for at most 5 cars failing

We use the formula for the cumulative binomial distribution probability, given by \(P(k≤n) = \sum_{i=0}^{k} C(n,i) * p^i * (1-p)^{n-i}\) for k=0 to 5. Here, n=15 (number of trials), p=0.3 (probability of failing) and C(n,i) is the binomial coefficient.

Calculate probability for between 5 and 10 cars failing

As above, we use the binomial distribution formula. However, this time we calculate the cumulative distribution from k=0 to 10, then subtract the cumulative distribution from k=0 to 4, thus giving the probability of 5 to 10 inclusive.

Calculate the mean and standard deviation

The mean value of a binomial distribution is \(np\) and the standard deviation is \(\sqrt{np(1-p)}\). Substituting the given values n=25 and p=0.3, we first find the mean and standard deviation for the cars failing the inspection. To find the values for those passing, we subtract from the total number of cars.

Calculate probability for number of passed cars within 1 standard deviation of the mean

Convert it into a standard normal variable Z=\((X-\mean)/\sd\). In this case, X is the number of cars that pass and is within one standard deviation, meaning X should be between \(\mean - \sd\) and \(\mean + \sd\). Use the standard normal distribution table to find the probability.