Answer:
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!!!!
