Question

Suppose that fuel efficiency for a particular model car under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this type is between 29 and \(31 \mathrm{mpg}\) ? b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg}\) ? c. If three cars of this model are randomly selected, what is the probability that all three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(c\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(c\) (i.e., \(P(x>c)=.95\) ).

Short answer

a) The probability that the efficiency is between 29 and 31 mpg is approximately 0.7967. b) It would be surprising to find a car with an efficiency less than 25 mpg. c) The probability that all three randomly selected cars have efficiencies exceeding 32 mpg is approximately 0.000107. d) The 'c' value such that 95% of all cars have efficiencies exceeding 'c' is approximately 28 mpg.

Step-by-step solution

Calculate Probability for Part a

To calculate the probability that a randomly selected car has a fuel efficiency between 29 and 31 mpg, first calculate the Z-scores associated with 29 and 31 mpg. The formula for Z score is \(Z = (X - \mu) / \sigma\). For 29 mpg \(Z_1 = (29 - 30) / 1.2 = -0.83\) and for 31 mpg \(Z_2 = (31 - 30) / 1.2 = 0.83\). Use a standard normal distribution (Z-table) to look up these Z-scores. The probability that the Z score is between -0.83 and 0.83 is 0.7967. Hence, the probability that a randomly selected car has an efficiency between 29 and 31 mpg is 0.7967.

Give judgement for Part b

We need to calculate the probability that a randomly selected car has an efficiency less than 25 mpg. First, calculate the Z score for 25 mpg, \( Z = (25 - 30) / 1.2 = -4.167\). Looking this Z score up in the standard normal distribution table gives a probability of less than 0.0001. Since this probability is extremely low, it would indeed be surprising to find a car with efficiency less than 25 mpg.

Calculate Probability for Part c

We need to calculate the probability that all three randomly selected cars have fuel efficiency exceeding 32 mpg. First, calculate the Z score for 32 mpg, \( Z = (32 - 30) / 1.2 = 1.67\). Looking up this Z score in the standard normal distribution table gives a probability of 0.0475. Since the cars are independently selected, the overall probability that all three cars have efficiencies exceeding 32 mpg is \(0.0475^3 = 0.000107\).

Find 'c' for Part d

We need to find a value 'c' such that 95% of all cars of this model have fuel efficiency exceeding 'c'. This means \(P(X > c) = 0.95\) or equivalently \(P(X \leq c) = 0.05\). The Z-score related to 0.05 in the lower tail of the standard normal distribution is approximately -1.645. Thus, we solve the formula \(c = \mu + Z \cdot \sigma\), with \(\mu = 30.0\), \(Z = -1.645\), and \(\sigma = 1.2\) to find 'c'. This gives \(c = 30 - 1.645 \cdot 1.2 = 28\).

Key concepts

Probability Calculation

Probability calculation is a fundamental concept in statistics that deals with the likelihood of an event occurring. When we're talking about normally distributed data, like the fuel efficiency of cars in our exercise, we calculate the probability of a value falling within a certain range.

Let's simplify this with an example from our exercise: If we want to know the probability of a car having fuel efficiency between 29 and 31 mpg, we use the normal distribution curve that represents all possible values of fuel efficiency and their probabilities. The area under the curve between these two points corresponds to our desired probability. In our case, that probability turned out to be 0.7967, which means there's about a 79.67% chance a randomly selected car will have an efficiency between 29 to 31 mpg. This calculation helps decision-makers, like car manufacturers or consumers, assess the performance of a car model in terms of fuel efficiency.

Z-score

The Z-score is a way of describing a value's relationship to the mean of a group of values, measured in terms of standard deviations. If a Z-score is 0, it indicates that the data point's score is identical to the mean score. A Z-score can be positive or negative, representing data points that are above or below the mean, respectively.

For example, in our fuel efficiency problem, we calculated the Z-score for a car with 29 mpg. Since the mean fuel efficiency is 30 mpg, and the standard deviation is 1.2 mpg, the Z-score of a car with 29 mpg was -0.83. This indicates that 29 mpg is 0.83 standard deviations below the mean. Z-scores turn individual data points into standardized values, making it possible to compare them to the overall distribution. This concept is critical for understanding how an individual data point relates to a larger set of data.

Standard Normal Distribution

The standard normal distribution is a special case of the normal distribution. It's a symmetrical distribution where most values cluster around the mean, and the probabilities for values can be found in standard Z-tables since the mean is always 0 and the standard deviation is 1. This concept is crucial because we can use it to find probabilities and percentiles for any normally distributed data set after converting our values using Z-scores.

To apply this to our car efficiency exercise, we transformed the mpg values into Z-scores and then used the standard normal distribution tables (Z-tables) to find corresponding probabilities. By standardizing different data sets to a common scale, we can use the standard normal distribution to compare different distributions effectively and efficiently.

Fuel Efficiency Statistics

Fuel efficiency statistics involve the collection and analysis of data regarding vehicles' fuel economy. These statistics help consumers make informed decisions, allow manufacturers to design fuel-efficient vehicles, and enable governments to set regulations on emissions and fuel economy standards.

In the exercise, we examined a normally distributed set of fuel efficiency data for a particular car model. Understanding the distribution of these statistics is important because it informs us about what kind of fuel efficiency we can expect from most cars of this model. Furthermore, by calculating probabilities like those in parts a, b, and c, we can gauge the likelihood of a car meeting or exceeding certain fuel efficiency thresholds. Finally, calculating a value like 'c' helps establish a benchmark or guarantee for minimum performance in fuel efficiency for a substantial percentage of cars. These statistics are powerful tools for comparison and expectation setting in the automobile industry.