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Suppose that the speed at which cars go on the freeway is normally distributed with mean 71 mph and standard deviation 9 miles per hour. Let X be the speed for a randomly selected car. Round all answers to two decimal places.

A. X ~ N( _____, _______ )
B. If one car is randomly chosen, find the probability that is traveling more than 70 mph.
C. If one of the cars is randomly chosen, find the probability that it is traveling between 73 and 79 mph.
D. 63% of all cars travel at least how fast on the freeway? ________ mph.

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Answer:

Explanation:

A. The speed of cars on the freeway is normally distributed with a mean (μ) of 71 mph and a standard deviation (σ) of 9 mph. So, you can express this as:

X ~ N(71, 9^2)

B. To find the probability that a randomly selected car is traveling more than 70 mph, you can use the standard normal distribution table or a calculator to find the z-score for 70 mph and then find the probability associated with that z-score being less than or equal to it. The formula for the z-score is:

\[z = \frac{X - μ}{σ}\]

Where X is the value you want to find the probability for (in this case, 70 mph), μ is the mean (71 mph), and σ is the standard deviation (9 mph).

\[z = \frac{70 - 71}{9} = -\frac{1}{9} \approx -0.11\]

Now, find the probability using the z-score:

P(X > 70) = 1 - P(X ≤ 70)

Using a standard normal distribution table or calculator, you can find P(X ≤ -0.11) and subtract it from 1 to get the answer.

C. To find the probability that a randomly selected car is traveling between 73 and 79 mph, you can calculate the z-scores for both 73 and 79 mph and find the probabilities associated with those z-scores. Then, subtract the probability associated with the lower z-score from the probability associated with the higher z-score.

\[z_1 = \frac{73 - 71}{9} = \frac{2}{9} \approx 0.22\]

\[z_2 = \frac{79 - 71}{9} = \frac{8}{9} \approx 0.89\]

Now, find the probabilities:

P(73 ≤ X ≤ 79) = P(0.22 ≤ Z ≤ 0.89)

You can use a standard normal distribution table or calculator to find these probabilities.

D. To find the speed at which 63% of all cars travel on the freeway, you need to find the z-score that corresponds to the 63rd percentile of the standard normal distribution. You can then use the z-score formula to find the speed associated with that z-score.

First, find the z-score corresponding to the 63rd percentile, denoted as z_(0.63). You can use a standard normal distribution table or calculator to find this value.

Once you have the z_(0.63) value, you can use it to find the speed (X) using the z-score formula:

\[z_(0.63) = \frac{X - μ}{σ}\]

Solve for X:

\[X = z_(0.63) * σ + μ\]

Substitute the values of z_(0.63), σ, and μ to calculate X.

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