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 = 3.7 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 button hyperlink to the SALT program that reads: Use SALT.
(a) an escape dune will move a total distance of more than 90 feet in 9 years

(b) an escape dune will move a total distance of less than 80 feet in 9 years

(c) an escape dune will move a total distance of between 80 and 90 feet in 9 years

email at dbrhnll8

(a) 90ft/9yr = 10 ft/year

That is 10/3.7 = 2.7𝜎 above the mean.
So what does your Z table say?

similarly for (b) and (c)

To calculate the probabilities in this scenario, we need to use the normal distribution and its properties. The formula for calculating probabilities in a normal distribution is given by:

P(X < a) = Φ((a - μ) / σ)

where P(X < a) represents the probability that a random variable X is less than a, Φ(z) is the standard normal cumulative distribution function, μ is the mean of the distribution, and σ is the standard deviation.

Let's calculate the probabilities for each of the given scenarios:

(a) The probability that an escape dune will move a total distance of more than 90 feet in 9 years.

In this case, we want to calculate P(X > 90), which is the probability that the random variable X is greater than 90.

Using the formula, we substitute a = 90, μ = 10, and σ = 3.7:

P(X > 90) = 1 - Φ((90 - 10) / 3.7)

Calculate (90 - 10) / 3.7, then find the corresponding value in the standard normal table or using a calculator that provides the cumulative distribution function. Subtract the value obtained from 1 to find the probability.

(b) The probability that an escape dune will move a total distance of less than 80 feet in 9 years.

In this case, we want to calculate P(X < 80), which is the probability that the random variable X is less than 80.

Using the formula, we substitute a = 80, μ = 10, and σ = 3.7:

P(X < 80) = Φ((80 - 10) / 3.7)

Calculate (80 - 10) / 3.7, then find the corresponding value in the standard normal table or using a calculator that provides the cumulative distribution function.

(c) The probability that an escape dune will move a total distance of between 80 and 90 feet in 9 years.

In this case, we want to calculate P(80 < X < 90), which is the probability that the random variable X is between 80 and 90.

Using the formula, we substitute a = 80, μ = 10, and σ = 3.7, yielding:

P(80 < X < 90) = Φ((90 - 10) / 3.7) - Φ((80 - 10) / 3.7)

Calculate [(90 - 10) / 3.7] and [(80 - 10) / 3.7], then find the corresponding values in the standard normal table or using a calculator that provides the cumulative distribution function. Subtract the second value from the first to find the probability.

Remember to round your answers to four decimal places, as instructed.