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natta225 [31]
3 years ago
12

New York City is the most expensive city in the United States for lodging. The room rate is $204 per night (USA Today, April 30,

2012). Assume that room ratesa mally distributed with a standard deviation of $55. a. What is the probability that a hotel room costs $225 or more per night? b. What is the probability that a hotel room costs less than $140 per night? c. What is the probability that a hotel room costs between $200 and $300 per night? d. What is the cost of the most expensive 20% of hotel rooms in New York City?
Mathematics
1 answer:
Sever21 [200]3 years ago
6 0

Answer:

a. 0.35197 or 35.20%; b. 0.1230 or 12.30%; c. 0.48784 or 48.78%; d. $250.20 or more.

Step-by-step explanation:

In general, we can solve this question using the <em>standard normal distribution</em>, whose values are valid for any <em>normally distributed data</em>, provided that they are previously transformed to <em>z-scores</em>. After having these z-scores, we can consult the table to finally obtain the probability associated with that value. Likewise, for a given probability, we can find, using the same table, the z-score associated to solve the value <em>x</em> of the equation for the formula of z-scores.

We know that the room rates are <em>normally distributed</em> with a <em>population mean</em> and a <em>population standard deviation</em> of (according to the cited source in the question):

\\ \mu = \$204 <em>(population mean)</em>

\\ \sigma = \$55 <em>(population standard deviation)</em>

A <em>z-score</em> is the needed value to consult the <em>standard normal table. </em>It is a transformation of the data so that we can consult this standard normal table to obtain the probabilities associated. The standard normal table has a mean  of 0 and a standard deviation of 1.

\\ z_{score}=\frac{x-\mu}{\sigma}

After having all this information, we can proceed as follows:

<h3>What is the probability that a hotel room costs $225 or more per night? </h3>

1. We need to calculate the z-score associated with x = $225.

\\ z_{score}=\frac{225-204}{55}

\\ z_{score}=0.381818

\\ z_{score}=0.38

We rounded the value to two decimals since the <em>cumulative standard normal table </em>(values for cumulative probabilities from negative infinity to the value x) to consult only have until two decimals for z values.

Then

2. For a z = 0.38, the corresponding probability is P(z<0.38) = 0.64803. But the question is asking for values greater than this value, then:

\\ P(z>038) = 1 - P(z (that is, the complement of the area)

\\ P(z>038) = 1 - 0.64803

\\ P(z>038) = 0.35197

So, the probability that a hotel room costs $225 or more per night is P(x>$225) = 0.35197 or 35.20%, approximately.

<h3>What is the probability that a hotel room costs less than $140 per night?</h3>

We follow a similar procedure as before, so:

\\ z_{score}=\frac{x-\mu}{\sigma}

\\ z_{score}=\frac{140-204}{55}

\\ z_{score}=\frac{140-204}{55}

\\ z_{score}= -1.163636 \approx -1.16

This value is below the mean (it has a negative sign). The standard normal tables does not have these values. However, we can find them subtracting the value of the probability obtained for z = 1.16 from 1, since the symmetry for normal distribution permits it. Then, the probability associated with z = -1.16 is:

\\ P(z

\\ P(z

\\ P(z

Then, the probability that a hotel room costs less than $140 per night is P(x<$140) = 0.1230 or 12.30%.

<h3>What is the probability that a hotel room costs between $200 and $300 per night?</h3>

\\ z_{score}=\frac{x-\mu}{\sigma}

<em>The z-score and probability for x = $200:</em>

\\ z_{score}=\frac{200-204}{55}

\\ z_{score}= -0.072727 \approx -0.07

\\ P(z

\\ P(z

\\ P(z

<em>The z-score and probability for x = $300:</em>

\\ z_{score}=\frac{300-204}{55}

\\ z_{score}=1.745454

\\ P(z

\\ P(z

\\ P(z

Then, the probability that a hotel room costs between $200 and $300 per night is 0.48784 or 48.78%.

<h3>What is the cost of the most expensive 20% of hotel rooms in New York City?</h3>

A way to solve this is as follows: we need to consult, using the cumulative standard normal table, the value for z such as the probability is 80%. This value is, approximately, z = 0.84. Then, solving the next equation for <em>x:</em>

\\ z_{score}=\frac{x-\mu}{\sigma}

\\ 0.84=\frac{x-204}{55}

\\ 0.84*55=x-204

\\ 0.84*55 + 204 =x

\\ x = 250.2

That is, the cost of the most expensive 20% of hotel rooms in New York City are of $250.20 or more.

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Answer: Choice B) Infinitely many solutions

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  • another solution: x = -12, y = 13/2, z = 10

=======================================================

Explanation:

Here's the starting original augmented matrix.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\-4 & 0 & -8 & -32\\\end{array}\right]

We'll multiply everything in row 3 (abbreviated R3) by the value -1/4 or -0.25, which will make that -4 in the first column turn into a 1.

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That notation says "multiply everything in R3 by -1/4, then replace the old R3 with the new corresponding values".

So we have this next step:

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\1 & 0 & 2 & 8\\\end{array}\right]\begin{array}{l}  \ \\\ \\(-1/4)*R3 \to R3\\\end{array}

Notice that the new R3 is perfectly identical to R1.

So we can subtract rows R1 and R3, and replace R3 with the result of nothing but 0's

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\\ \\R3-R1 \to R3\\\end{array}

Whenever you get an entire row of 0's, it <u>always</u> means there are infinitely many solutions.

-------------------

Now let's handle the second row. That 5 needs to turn into a 0. We can multiply R1 by 5, and subtract that from R2.

So we need to compute 5*R1-R2 and have that replace R2.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\0 & 1 & -1 & -7/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\5*R1-R2 \to R2\ \\\ \\\end{array}

Notice that in the third column of R2, we have 9-5*2 = 9-10 = -1. So we have -1 replace the 9. In the fourth column of R2, we have 73/2 - 5*8 = -7/2. So the -7/2 replaces the 73/2.

--------------------

At this point, the augmented matrix is in RREF form. RREF stands for Reduced Row Echelon Form. It seems a bit odd that the "F" of "RREF" stands for "form" even though we say "form" right after "RREF", but I digress.

Because the matrix is in RREF form, this means R1 and R2 lead to these equations:

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which simplify to

R1: x+2z = 8\\R2: y-z = -7/2

Let's get the z terms to each side like so:

x+2z = 8\\x = -2z+8\\\text{ and }\\y-z = -7/2\\y = z-7/2\\

Therefore, all of the solutions are of the form (x,y,z) = (-2z+8, z-7/2, z) where z is any real number.

If z is allowed to be any real number, then we can simply pick any number we want to replace it. We consider z to be the "free variable", in that it's free to be whatever it wants. The values of x and y will depend on what we pick for z.

So the concept of "infinitely many solutions" doesn't exactly mean we can pick just <em>any</em> triple for x,y,z (admittedly it would be nice to randomly pick any 3 numbers off the top of my head and be done right away). Instead, we can pick anything we want for z, and whatever we picked, will directly determine x and y. The x and y are locked into place so to speak.

Let's say we picked z = 0.

That would lead to...

x = -2z+8\\x = -2(0)+8\\x = 8\\\text{ and }\\y = z-7/2\\y = 0-7/2\\y = -7/2\\

So z = 0 would lead to x = 8 and y = -7/2

Rearranging the items in alphabetical order gets us:

x = 8, y = -7/2, z = 0

We have one solution of (x,y,z) = (8, -7/2, 0)

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x = -2z+8\\x = -2(10)+8\\x = -12\\\text{ and }\\y = z-7/2\\y = 10-7/2\\y = 13/2\\

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Another solution is (x,y,z) = (-12, 13/2, 10)

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To verify each solution, you'll need to plug them back into the original equations formed by the original augmented matrix. After simplifying, you should get the same thing on both sides.

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