Question

Speeding on the I-5. Suppose the distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 73 miles/hour and a standard deviation of 4.65 miles/hour. Round all answers to four decimal places. What proportion of passenger vehicles travel slower than 72 miles/hour? What proportion of passenger vehicles travel between 66 and 73 miles/hour? How fast do the fastest 6% of passenger vehicles travel? miles/hour Suppose the speed limit on this stretch of the I-5 is 70 miles/hour. Approximately what proportion of the passenger vehicles travel above the speed limit on this stretch of the I-5?

Answer 1

Solution :

Given that ,

mean = = 73

standard deviation = = 4.65

(a)

P(x < 72) = P((x - ) / < (72 - 73) / 4.65)

= P(z < 0.215)

= 0.5851

Probability = 0.25851

.(b)

P(66 < x < 73) = P((66 - 73)/ 4.65) < (x - ) /  < (73 - 73) / 4.65) )

= P( -1.5054 < z < 0)

= P(z < 0) - P(z < -1.5054)

= 0.5 - 0.0661

= 0.4339

Probability = 0.4339

(c)

P(Z > z) = 6%

1 - P(Z < z) = 0.06

P(Z < z) = 1 - 0.06

P(Z < 1.555) = 0.94

z = 1.555

Using z-score formula,

x = z * +

x = 1.555 * 4.65 + 73 = 80.23

Speed = 80.23 miles

(d)

P(x > 70) = 1 - P(x < 70)

= 1 - P((x - ) / < (70 - 73) / 4.65)

= 1 - P(z < 0.6452)

= 1 - 0.7406   

= 0.2594

Probability = 0.2594