What is the range of the function graphed below?

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,

Sever21 [200]

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 standard normal distribution, whose values are valid for any normally distributed data, provided that they are previously transformed to z-scores. 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 x of the equation for the formula of z-scores.

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

$\\ \mu = \$204$ (population mean)

$\\ \sigma = \$55$ (population standard deviation)

A z-score is the needed value to consult the standard normal table. 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 cumulative standard normal table (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:

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

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

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

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:

$\\ P(z$

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>

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

The z-score and probability for x = $200:

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

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

$\\ P(z$

The z-score and probability for x = $300:

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

$\\ z_{score}=1.745454$

$\\ P(z$

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 x:

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

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