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A histogram is a graphical representation that organizes a group of data points into user-specified ranges. In this context, we use it to show the probability distribution of random variable \( r \), representing the number of vehicles that tailgate before passing.
When dealing with a binomial distribution, like in this problem where each car has a 35% probability of tailgating, a histogram can effectively visualize these probabilities. Here’s how you do it:

• First, determine the number of trials (which is 12, the number of cars) and the probability of a car tailgating (\( p = 0.35 \)).
• Use the binomial probability formula: \[ P(r) = \binom{12}{r} \times (0.35)^r \times (0.65)^{12-r} \]
• Calculate this for each possible value of \( r \) (from 0 up to 12).
• Plot these probabilities as bars on the histogram, with \( r \) on the x-axis and \( P(r) \) on the y-axis.

The resulting graph helps us easily see which outcomes (values of \( r \)) are more likely than others. This makes it clear how often we can expect a certain number of cars to tailgate as they pass.

Expected value is a key concept in statistics that provides a means of predicting the average outcome of a random process over an infinite number of trials. In a binomial distribution, it gives us an estimate of the number of times an event will occur.
For our situation with tailgating drivers, the expected number of cars that will tailgate can be calculated using this formula: \[E(r) = n \times p\]
Where \( n \) is the total number of trials and \( p \) is the probability of success for each trial. With \( n = 12 \) cars and a tailgating probability of \( p = 0.35 \), we find:\[E(r) = 12 \times 0.35 = 4.2\]
This result of 4.2 suggests that, on average, a bit more than 4 cars out of 12 will tailgate. While you can’t have a fraction of a car tailgating in practice, this figure gives a useful average or expectation across many repeated scenarios of similar situations.

Standard deviation is a measure that is used to quantify the amount of variation or dispersion of a set of values. In simple terms, it tells us about how spread out the results can be. For a binomial distribution such as our example, the standard deviation helps us understand how much individual observations of tailgating cars can stray from the expected average.
The formula for the standard deviation in a binomial distribution is:
\[ \sigma = \sqrt{n \times p \times (1-p)} \]
Substituting our values, where \( n = 12 \) and \( p = 0.35 \), we compute:\[ \sigma = \sqrt{12 \times 0.35 \times 0.65} \approx 1.67 \]
This value of approximately 1.67 means that the number of tailgating cars can typically deviate by about 1.67 cars from the expected value of 4.2 across multiple observed trials. Thus, if you imagine repeating this exercise of being passed by 12 cars many times, this standard deviation guides us on the variability we might observe in the number of tailgaters each time.