Question

A gasoline tank for a certain car is designed to hold 15 gal of gas. Suppose that the variable \(x=\) actual capacity of a randomly selected tank has a distribution that is well approximated by a normal curve with mean \(15.0 \mathrm{gal}\) and standard deviation \(0.1\) gal. a. What is the probability that a randomly selected tank will hold at most \(14.8\) gal? b. What is the probability that a randomly selected tank will hold between \(14.7\) and \(15.1\) gal? c. If two such tanks are independently selected, what is the probability that both hold at most 15 gal?

Short answer

a. The probability that a tank will hold at most 14.8 gal is approximately 0.02275. b. The probability that a tank will hold between 14.7 and 15.1 gal is approximately 0.84. c. The probability that both tanks hold at most 15 gal is 0.25.

Step-by-step solution

- Understanding the distribution

The problem informs us that the capacity of gas tanks follows a normal distribution with a mean (\(\mu\)) of 15.0 gal and a standard deviation (\(\sigma\)) of 0.1 gal. When a distribution is normal, most of the data falls close to the mean, and the spread of data is symmetrical about the mean. In this case, most tanks will hold close to 15 gallons, with few holding significantly more or less.

Step 2(a) - Calculating the probability of a tank holding at most 14.8 gal

To calculate this, we first need to find the Z-score for 14.8 gal. The Z-score measures how many standard deviations an element is from the mean, and is given by Z = (X - \(\mu\)) / \(\sigma\). Substituting the given values, we get Z = (14.8 - 15) / 0.1 = -2. Now, we look up this Z-score in a standard normal distribution table or use the cumulative distribution function of a standard normal distribution to find out the probability, which gives us approximately 0.02275.

Step 3(b) - Calculating the probability of a tank holding between 14.7 and 15.1 gal

We need to find the Z-scores for 14.7 and 15.1. Using the same formula, we get Z = (14.7 - 15) / 0.1 = -3 for 14.7 gal and Z = (15.1 - 15) / 0.1 = 1 for 15.1 gal. Now, we need to find the area under the normal curve between these two Z-scores. That area represents the probability we are looking for. Using the standard normal distribution, we get Prob(14.7 < X < 15.1) = Prob(-3 < Z < 1) = Prob(Z < 1) - Prob(Z < -3) = 0.84134 - 0.00135 = 0.84 approximately.

Step 4(c) - Calculating probability of both tanks holding at most 15 gal

The Z-score for 15 is Z = (15 - 15) / 0.1 = 0. Now, we need to find the probability of one tank holding at most 15 gallons, and then square it, since the two events are independent and we need both of them to happen. Prob(X ≤ 15) = Prob(Z ≤ 0) = 0.5. Hence, the probability that both tanks hold at most 15 gallons is (0.5)² = 0.25.