3.12 Speeding on the I-5, Part I. 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.47 (a) What percent of passenger vehicles travel slower than 80 miles/hour(b) What percent of passenger vehicles travel between 60 and 80 miles/hour? (c) How fast to do the fastest 5% of passenger vehicles travel? (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.

a) 93.943% = 93.9%

b) 93.528% = 93.5%

c) Speed of the fastest 5% ≥ 80.5 miles/hour

d) 29.46% = 29.5%

Step-by-step explanation:

Mean, xbar = 72.6 miles/hour.

standard deviation, σ = 4.78 miles/hour

For each of the questions, we'll need to normalize the speeds.

a) The standardized score for 80 miles/hour is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ = (80 - 72.6)/4.78 = 1.55

To determine the probability of a car having speed less than 80 miles/hour, P(x < 80) = P(z < 1.55)

We'll use data from the normal probability table for these probabilities

P(x < 80) = P(z < 1.55) = 1 - P(z ≥ 1.55) = 1 - P(z ≤ -1.55) = 1 - 0.06057 = 0.93943

b) percent of passenger vehicles travel between 60 and 80 miles/hour.

60 miles/hour standardized = (60 - 72.6)/4.78 = -2.64

We'll use data from the normal probability table for these probabilities

P(60 < x < 80) = P(-2.64 < z < 1.55) = P(z ≤ 1.55) - P(z ≤ -2.64) = 0.93943 - 0.00415 = 0.93528

c) How fast to do the fastest 5% of passenger vehicles travel?

We'll use data from the normal probability table for these probabilities

Top 5% corresponds to a z-score of 1.65. P(z ≥ 1.65) = 0.95053

1.65 = (x - 72.6)/4.78

x = 80.487 miles/hour = 80.5 miles/hour.

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.

70 miles/hour, standardized = (70 - 72.6)/4.78 = 0.54

P(x > 70) = P(z > 0.54) = 1 - P(z ≤ 0.54) = 1 - 0.7054 = 0.2946.

Hope this helps

74.25 gallons of milk per day

Step-by-step explanation:

6.75 gallons of milk/cow×11 cows=74.25gallons of milk per day