Question
Asked Dec 4, 2019
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30. The speeds (in mph) of motor vehicles on a certain stretch of Route 3A as
clocked at a certain place along the highway are normally distributed with a
mean of 64.2 mph and a standard deviation of 8.44 mph. What is the
probability that a motor vehicle selected at random is traveling at...
(A) More than 65 mph?
(B) Less than 60 mph?
o3 Xe 00/9 OH
007
(C) Between 65 and 70 mph?
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30. The speeds (in mph) of motor vehicles on a certain stretch of Route 3A as clocked at a certain place along the highway are normally distributed with a mean of 64.2 mph and a standard deviation of 8.44 mph. What is the probability that a motor vehicle selected at random is traveling at... (A) More than 65 mph? (B) Less than 60 mph? o3 Xe 00/9 OH 007 (C) Between 65 and 70 mph?

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Expert Answer

Step 1

Mean (µ) = 64.2

Standard Deviation (σ) = 8.44

Probability that a motor vehicle selected at random is travelling at

A) More than 65 mph is given by 

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P(X > 65) Changing into standard normal variate 65 – 64.2 = P(z > 0.0948) P(z> 8.44 P(z > 0.0948) = 1 – P(z < 0.0948) = 1 – 0.5378 = 0.4622 (P(z < 0.0948) = 0.5378(From Excel NORM. S. DIST(0.0948, TRUE)) B) Less than 60 mph is given by P(X < 60) Changing into standard normal variate 60 – 64.2 P(z< = P(z < -0.4976) 8.44 (P(z < -0.4976) = 0.3094(From Excel NORM. S. DIST(-0.4976, TRUE))

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Step 2

C) Between 65 and 75 mph is gi...

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P(65 < X < 75) = P(X < 75) – P(X < 65) Changing into standard normal variate 65 – 64.2 65-µ 75 — и ")- 75 – 64.2 -)-P(25 - P(2= 8.44 8.44 = P(z < 1.2796) – P(z < 0.0948) NORM.S. DIST(1.2796, TRUE) (P(z < 1.2796) = 0.8997 From excel (P(z < 0.0948) = 0.5378(From Excel NORM. S. DIST(0.0948, TRUE)) = 0.8997 – 0.5378 = 0.3619

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