Question

It's true — sand dunes in Colorado rival sand dunes of the Great Sahara Desert! The highest dunes at Great Sand Dunes National Monument can exceed the highest dunes in the Great Sahara, extending over 700 feet in height. However, like all sand dunes, they tend to move around in the wind. This can cause a bit of trouble for temporary structures located near the "escaping" dunes. Roads, parking lots, campgrounds, small buildings, trees, and other vegetation are destroyed when a sand dune moves in and takes over. Such dunes are called "escape dunes" in the sense that they move out of the main body of sand dunes and, by the force of nature (prevailing winds), take over whatever space they choose to occupy. In most cases, dune movement does not occur quickly. An escape dune can take years to relocate itself. Just how fast does an escape dune move? 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 = 10 feet per year and σ = 3.8 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 9 years .5000 Correct: Your answer is correct. (b) an escape dune will move a total distance of less than 80 feet in 9 years .3850 Incorrect: Your answer is incorrect. (c) an escape dune will move a total distance of between 80 and 90 feet in 9 years

Answer 1

Sol:

Since x has a normal distribution and the mean and the standard deviation is given for 1 year, therefore for 9 years.

= 10 * 9= 90 ft/ 10years

and

= 3.8* sqrt(9) = 11.4ft/10years

To find the probability, we need to find the Z scores first.

Z = (X - µ)/ [σ/√n]. Since n = 1, Z = (X - µ)/ σ

(a).

For P (X > 90) = 1 - P (X < 90), as the normal tables give us the left tailed probability only.

For P( X < 90)

Z = (90 – 90)/11.4

Z = 0

The probability for P(X < 90) from the normal distribution tables is = 0500

Therefore the required probability = 1 – 0.5000

Therefore the required probability = 0.500

(b).

For P( X < 80)

Z = (80 – 90)/11.4

Z = -0.877

The required probability from the normal distribution tables is = 0.4761

(c).

For P (80 < X < 90) = P(X < 90) – P(X < 80)

From (a) P(X < 90) = 0.500 and from (b) P(X < 80) = 0.4761

Therefore the required probability = 0.500 – 0.4761

Therefore the required probability = 0.0239

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