Question

The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 75 miles/hour and a standard deviation of 6.1 miles/hour.

(a) What proportion of passenger vehicles travel slower than 77 miles/hour?

(b) What proportion of passenger vehicles travel between 65 and 81 miles/hour?

(c) How fast do the fastest 15% of passenger vehicles travel? miles/hour

(d) Find a value k so that 50% of passanger vehicles travel at speeds within k miles/hour of 75mph. 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

This is a normal distribution question with

a) P(x < 77.0)=?
The z-score at x = 77.0 is,

z = 0.3279
This implies that
P(x < 77.0) = P(z < 0.3279) = 0.6285
b) P(65.0 < x < 81.0)=?

This implies that
P(65.0 < x < 81.0) = P(-1.6393 < z < 0.9836) = P(Z < 0.9836) - P(Z < -1.6393)
P(65.0 < x < 81.0) = 0.8373 - 0.0505
P(65.0 < x < 81.0) = 0.7868
c) Given in the question
P(X < x) = 0.85
This implies that
P(Z < 1.0364) = 0.85
With the help of formula for z, we can say that

x = 81.3222
d) Given in the question
P(X < x) = 0.75 (at 75% the upper limit of range so that 50% of passanger vehicles travel at speeds within k miles/hour of 75mph)
This implies that
P(Z < 0.6745) = 0.75
With the help of formula for z, we can say that

x = 79.1144
k = 79.1144 - 75 = 4.1144
e) x = 70
P(x > 70.0)=?
The z-score at x = 70.0 is,

z = -0.8197
This implies that
P(x > 70.0) = P(z > -0.8197) = 1 - 0.2062
P(x > 70.0) = 0.7938
PS: you have to refer z score table to find the final probabilities.
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