• anonymous
Let x be a random variable representing movement (in feet per year) of such sand dunes (measured from the crest of the dune). Let us assume that x has a normal distribution with μ = 11 feet per year and σ = 2 feet per year. Under the influence of prevailing wind patterns, what is the probability of each of the following? (Round your answers to four decimal places.) (a) an escape dune will move a total distance of more than 90 feet in 7 years
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • chestercat
I got my questions answered at in under 10 minutes. Go to now for free help!
  • kropot72
The first step is to convert the distribution to one that applies to distances over a period of 7 years. The 7 year distribution has \[\large \mu=11\times7=77\ feet:\ \ \ \ \sigma=2\times7=14\ feet\] The z-score for a distance of 90 feet is found from: \[\large z=\frac{X-\mu}{\sigma}=\frac{90-77}{14}=0.9286\] Now you need to refer to a standard normal distribution table to find the required probability.

Looking for something else?

Not the answer you are looking for? Search for more explanations.