Question

1.)

Bringing all of section 6.2 together...

The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-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 I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.
% (round to two decimal places)

2.)

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 230-cm and a standard deviation of 2.4-cm. Suppose a rod is chosen at random from all the rods produced by the company. There is a 39% probability that the rod is longer than:

Enter your answer as a number accurate to 1 decimal place.

Answer 1

1)a) P(X < 80)

       = P((X - )/ < (80 - )/)

       = P(Z < (80 - 72.6)/4.78)

       = P(Z < 1.55)

       = 0.9394 = 93.94%

b) P(60 < X < 80)

= P((60 - )/ < (X - )/ < (80 - )/)

= P((60 - 72.6)/4.78 < Z < (80 - 72.6)/4.78)

= P(-2.64 < Z < 1.55)

= P(Z < 1.55) - P(Z < -2.64)

= 0.9394 - 0.0041

= 0.9353

= 93.53%

c) P(X > x) = 0.05

or, P((X - )/ > (x - )/) = 0.05

or, P(Z > (x - 72.6)/4.78) = 0.05

or, P(Z < (x - 72.6)/4.78) = 0.95

or, (x - 72.6)/4.78 = 1.645

or, x = 1.645 * 4.78 + 72.6

or, x = 80.46

d) P(X > 70)

= P((X - )/ > (70 - )/)

= P(Z > (70 - 72.6)/4.78)

= 1 - P(Z < (70 - 72.6)/4.78)

= 1 - P(Z < -0.54)

= 1 - 0.2946

= 0.7054

= 70.54%

2) P(X > x) = 0.39

or, P((X - )/ > (x - )/) = 0.39

or, P(Z > (x - 230)/2.4) = 0.39

or, P(Z < (x - 230)/2.4) = 0.61

or, (x - 230)/2.4 = 0.28

or, x = 0.28 * 2.4 + 230

or, x = 230.7