Question

The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 74.9 miles/hour and a standard deviation of 6.5 miles/hour.

(a) What proportion of passenger vehicles travel slower than 78 miles/hour?
(b) What proportion of passenger vehicles travel between 57 and 83 miles/hour?
(c) How fast do the fastest 10% of passenger vehicles travel? miles/hour
(d) Find a value k so that 45% of passanger vehicles travel at speeds within k miles/hour of 74.9mph. k=
(e) 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.

Answer 1

A) P(X < 78)

= P((X - )/ < (78 - )/)

= P(Z < (78 - 74.9)/6.5)

= P(Z < 0.48)

= 0.6844

B) P(57 < X < 83)

= P((57 - )/ < (X - )/ < (83 - )/)

= P((57 - 74.9)/6.5 < Z < (83 - 74.9)/6.5)

= P(-2.75 < Z < 1.25)

= P(Z < 1.25) - P(Z < -2.75)

= 0.8944 - 0.0030

= 0.8914

C) P(X > x) = 0.1

Or, P((X - )/ > (x - )/) = 0.1

Or P(Z > (x - 74.9)/6.5) = 0.1

Or, P(Z < (x - 74.9)/6.5) = 0.9

Or, (x - 74.9)/6.5 = 1.28

Or, x = 1.28 * 6.5 + 74.9

Or, x = 83.22

d) P(74.9 - k < X < 74.9 + k) = 0.45

Or, P((74.9 - k - )/ < (X - )/ < (74.9 + k - )/) = 0.45

P((74.9 - k - 74.9)/6.5 < Z < (74.9 + k - 74.9)/6.5) = 0.45

Or, P(-k/6.5 < Z < k/6.5) = 0.45

Or, P(Z < k/6.5) - P(Z < -k/6.5) = 0.45

Or, P(Z < k/6.5) - (1 - P(Z < k/6.5)) = 0.45

Or, P(Z < k/6.5) - 1 + P(Z < k/6.5) = 0.45

Or, 2P(Z < k/6.5) = 1.45

Or, P(Z < k/6.5) = 0.725

Or k/6.5 = 0.60

Or, k = 3.9

e) P(X > 70)

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

= P(Z > (70 - 74.9)/6.5)

= P(Z > -0.75)

= 1 - P(Z < -0.75)

= 1 - 0.2266

= 0.7734