Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation

Question

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation of 8 mph. a. The current speed limit is 65 mph. What is the proportion of vehicles less than or equal to the speed limit? b. What proportion of the vehicles would be going less than 50 mph? c. A new speed limit will be initiated such that approximately 10% of vehicles will be over the speed limit. What is the new speed limit based on this criterion? d. In what way do you think the actual distribution of speeds differs from a normal distribution?

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Adalynn 2 weeks 2021-11-08T14:26:56+00:00 1 Answer 0 views 0

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    2021-11-08T14:28:24+00:00

    Answer:

    a.  22,7% of the vehicles are going less than or equal to 65 mph.

    b. 0,4% of the vehicles are going less than 50 mph.

    c. if the speed limit would be 81.25 mph, aproximately 10% of vwhicles would be over that speed.

    Step-by-step explanation:

    \mu=71\\\sigma=8

    Part A.

    The portion of vehicles with speed less than or equal to 65 mph, could be estimate by the probability of X≤65. If the speed follow a normal distribution, the probability is:

    P(X\leq 65)=P(Z\leq \frac{X-\mu}{\sigma})\\P(X\leq 65)=P(Z\leq \frac{65-71}{8})\\P(X\leq 65)=P(Z\leq -0.75)

    According to normal distribution table, P(Z\leq -0.75)=0.227

    It means that 22,7% of the vehicles are going less than or equal to 65 mph.

    Part B.

    The portion of vehicles with speed less than 50 mph, could be estimate by the probability of X<50. If the speed follow a normal distribution, the probability is:

    P(X<50)=P(Z< \frac{X-\mu}{\sigma})\\P(X< 50)=P(Z< \frac{50-71}{8})\\P(X<50)=P(Z< -2.625)

    According to normal distribution table, P(Z< -2.625)=0.004

    It means that 0,4% of the vehicles are going less than 50 mph.

    Part C.

    1. Find Z-value

    The portion of vehicles that will be over the speed limit, could be expressed like:

    P(Z>SL)=0.1

    When, SL is the speed limit in a santdard normal distribution.

    P(Z>SL)=1- P(Z\leq SL)=0.1\\P(Z\leq SL)=0.9

    Using the normal distribution table, we will find the Z- value corresponding to probability of 0.9

    We find that Z- value for p=0.9 is 1.282

    2. Transform Z in X

    It’s necessary transforming Z-value in X, by the equation:

    Z=\frac{X-\mu}{\sigma}\\X=\sigma Z+\mu

    Then:

    X=(8)(1.282)+71\\X=81.25

    It means that if the speed limit would be 81.25 mph, aproximately 10% of vwhicles would be over that speed.

    Part D.

    The distribution doesn’t seem to be symmetric, the values are not distributed symmetrically around the mean.

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