The digits of a two-digit number differ by 2. If the digits are interchanged and the resulting number is added to the original n
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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):
<em>(population mean)</em>
<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.
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.
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:
(that is, the complement of the area)
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:
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:
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><em>The z-score and probability for x = $200:</em>
<em>The z-score and probability for x = $300:</em>
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>
That is, the cost of the most expensive 20% of hotel rooms in New York City are of $250.20 or more.