assume that the speed of automobiles on an expressway during rush hour is normally distributed with a mean of 62 mph and standard deviation of 5 mph. What percent of cars are traveling faster than 62 mph?
How would I work this problem out?? Thanks for any and all help anyone can give
What percent is traveling faster then the mean? By definition, 50 percent are faster, 50 percent slower than the mean.
Thanks Bob. Can you show me how you came up with your answer. That is what I am struggling with Thanks
In a normal distribution, the mean, median and mode have essentially the same value. Therefore, if the mean = 62, the median = 62. How is the median defined?
To determine the percentage of cars traveling faster than 62 mph, we need to calculate the area under the normal distribution curve to the right of 62 mph.
Step 1: Standardize the value of 62 mph using the z-score formula:
z = (x - μ) / σ
Where:
x = 62 mph (mean speed)
μ = mean of the distribution (62 mph)
σ = standard deviation of the distribution (5 mph)
Substituting the values, we have:
z = (62 - 62) / 5
z = 0
Step 2: Look up the area corresponding to the z-score of 0 in the standard normal distribution table. The table usually provides the area to the left of the z-score.
Since our interest lies to the right of our z-score, we can subtract the value obtained in Step 2 from 1 to get the area to the right.
Step 3: Calculate the percentage:
Percentage = Area to the right x 100
By following these steps, we can find the percentage of cars traveling faster than 62 mph.