Question

Bringing all of section 6.2 together... The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (1-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. (a) What percent of passenger vehicles travel slower than 80 miles/hour? % (round to two decimal places) (b) What percent of passenger vehicles travel between 60 and 80 miles/hour? % (round to two decimal places) (c) How fast do the fastest 5% of passenger vehicles travel? mph (round to two decimal places) (d) The speed limit on this stretch of the 1-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the 1-5. % (round to two decimal places)

Answer 1

Eliven Mean (u) = 72.6 miles Ela / m So C) = 4.78 miler Ihr ca). p(x< 80) = P(X-P < 80-7216 4.78 = 7.4 4.78 = P(z. z< 1.55) = 0.9394292 (from standard normal table) = 93.94%

(6) Xal 80-7216 2 (602*480) = 1(604:48 < 4.78 = -12.6 4.78 < Z < 7.4 4.78 - Р (-2.64 2.64 <Z < 1.55) P(Z<1.55) - P(Z < -2.64) = P = 0.9394292 - 0.0041453 (from standard normal table) 0.9352839 = 93.53%

(C). P(Xxx) 5% 3 0.05 1- P(x<a) 0.05 P(x<x) 0.95 from standard normal table P(Z <z) c0.95 By using z alpha table P (Z < 1.645) = 0.95 2 -1.645

We know that, 4 X-р X-72:6 Z - 1.645 19 o 4.78 x = 1.645 (4.88) + 72-6 = 7.8631 +72-6 = 80.463) X = 80.46 mph d). P(x> 40) = P(xonas 70-72-6 4.78 EP (z 2 > -2.6 4.78

15 c P (2> -0.54) El- P(Z <-0.54) C1-0.2945985 (from standard normal table) 0.7054014 - 70.54%

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