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Suppose that the speed at which cars go on the freeway is normally distributed with mean 65 mph and
standard deviation 6 miles per hour. Let X be the speed for a randomly selected car
b. If one car is randomly chosen, find the probability that it is traveling more than 63 mph.

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Answer:

B) the probability that is traveling more than 63 mph is 0.6293 (or 62.93%)

Step-by-step explanation:

Given:

  • Normally Distributed
  • Mean (μ) = 65 mph
  • Standard Deviation (σ) = 6 miles per hour

Finding the Probability:

If one car is randomly chosen, we want the probability that is traveling more than 63 mph is,

P(X > 63)

To find the value of z,

z = x - μ / σ

  • z is the standard score
  • x is the observed value
  • μ is the mean of the sample
  • σ is the standard deviation of the sample

z = 63 - 65 / 6

z = -2 / 6

z = -1 / 3 which is approximately -0.33

Using Z table (attached below):

  • z = -0.33

to find this on the table

  • on the vertical side under z go to -0.3
  • on the horizontal next to z, go to .03

The area under the curve is 0.3707

P(z > 63) = 1 - P(z < 63)

               = 1 - 0.3707

               = 0.6293

Hence the probability that is traveling more than 63 mph is 0.6293

Learn more Probability from a similar example: https://brainly.com/question/15565069

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