Math / Data & Statistics

QuestionA hotel manager assumes that $18 \%$ of the hotel rooms are booked. If the manager is correct, what is the probability that the proportion of rooms booked in a sample of 480 rooms would be less than $15 \%$ ? Round your answer to four decimal places.
Answer Tables Keypad
How to enter your answer (opens in new window) Keyboard Shortcuts

Studdy Solution

STEP 1

What is this asking? If 18% of hotel rooms should be booked, what's the chance that less than 15% are booked if we check 480 rooms? Watch out! Don't mix up percentages and proportions!
Also, remember we're dealing with a sample of rooms, not the entire hotel.

STEP 2

1. Set up the problem
2. Calculate the standard deviation
3. Calculate the z-score
4. Find the probability

STEP 3

Alright, let's define some important stuff!
We've got our population proportion, which is the assumed booking rate of $p = 0.18$ .
We're looking at a sample size of $n = 480$ rooms.
We want to know the probability that the sample proportion $\hat{p}$ is less than $0.15$ .

STEP 4

The standard deviation of the sample proportion tells us how spread out our sample results are likely to be.
The formula is: $\sigma_{\hat{p}} = \sqrt{\frac{p \cdot (1 - p)}{n}}$

STEP 5

Let's plug in our population proportion $p = 0.18$ and our sample size $n = 480$ : $\sigma_{\hat{p}} = \sqrt{\frac{0.18 \cdot (1 - 0.18)}{480}} = \sqrt{\frac{0.18 \cdot 0.82}{480}} = \sqrt{\frac{0.1476}{480}} \approx \sqrt{0.0003075} \approx 0.0175$ So, our standard deviation is approximately $0.0175$ .

STEP 6

The z-score tells us how many standard deviations away our target sample proportion ( $0.15$ ) is from the expected population proportion ( $0.18$ ).
The formula is: $z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}$

STEP 7

Plugging in our values: $z = \frac{0.15 - 0.18}{0.0175} = \frac{-0.03}{0.0175} \approx -1.71$ Our z-score is approximately $-1.71$ .

STEP 8

Now, we look up our z-score of $-1.71$ in a standard normal distribution table (or use a calculator).
This tells us the probability of getting a z-score less than $-1.71$ .

STEP 9

The table or calculator gives us a probability of approximately $0.0436$ .

STEP 10

The probability that the proportion of rooms booked in a sample of 480 rooms would be less than 15% is approximately 0.0436.

Was this helpful?

Studdy Solution

STEP 1

What is this asking? If 18% of hotel rooms should be booked, what's the chance that less than 15% are booked if we check 480 rooms? Watch out! Don't mix up percentages and proportions!
Also, remember we're dealing with a sample of rooms, not the entire hotel.

STEP 2

1. Set up the problem
2. Calculate the standard deviation
3. Calculate the z-score
4. Find the probability

STEP 3

Alright, let's define some important stuff!
We've got our population proportion, which is the assumed booking rate of $p = 0.18$ .
We're looking at a sample size of $n = 480$ rooms.
We want to know the probability that the sample proportion $\hat{p}$ is less than $0.15$ .

STEP 4

The standard deviation of the sample proportion tells us how spread out our sample results are likely to be.
The formula is: $\sigma_{\hat{p}} = \sqrt{\frac{p \cdot (1 - p)}{n}}$

STEP 5

STEP 6

The z-score tells us how many standard deviations away our target sample proportion ( $0.15$ ) is from the expected population proportion ( $0.18$ ).
The formula is: $z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}$

STEP 7

Plugging in our values: $z = \frac{0.15 - 0.18}{0.0175} = \frac{-0.03}{0.0175} \approx -1.71$ Our z-score is approximately $-1.71$ .

STEP 8

Now, we look up our z-score of $-1.71$ in a standard normal distribution table (or use a calculator).
This tells us the probability of getting a z-score less than $-1.71$ .

STEP 9

The table or calculator gives us a probability of approximately $0.0436$ .

STEP 10

The probability that the proportion of rooms booked in a sample of 480 rooms would be less than 15% is approximately 0.0436.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

What is this asking? If 18% of hotel rooms *should* be booked, what's the chance that *less than* 15% are booked if we check 480 rooms? Watch out! Don't mix up percentages and proportions!Also, remember we're dealing with a *sample* of rooms, not the entire hotel.

STEP 2

1. Set up the problem2. Calculate the standard deviation3. Calculate the z-score4. Find the probability

STEP 3

STEP 4

The **standard deviation** of the sample proportion tells us how spread out our sample results are likely to be.The formula is: σp^=p⋅(1−p)n \sigma_{\hat{p}} = \sqrt{\frac{p \cdot (1 - p)}{n}} σp^​​=np⋅(1−p)​​

STEP 5

STEP 6

The **z-score** tells us how many standard deviations away our **target sample proportion** (0.150.150.15) is from the **expected population proportion** (0.180.180.18).The formula is: z=p^−pσp^ z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} z=σp^​​p^​−p​

STEP 7

Plugging in our values: z=0.15−0.180.0175=−0.030.0175≈−1.71 z = \frac{0.15 - 0.18}{0.0175} = \frac{-0.03}{0.0175} \approx -1.71 z=0.01750.15−0.18​=0.0175−0.03​≈−1.71 Our **z-score** is approximately −1.71-1.71−1.71.

STEP 8

Now, we look up our **z-score** of −1.71-1.71−1.71 in a **standard normal distribution table** (or use a calculator).This tells us the probability of getting a z-score less than −1.71-1.71−1.71.

STEP 9

The table or calculator gives us a probability of approximately 0.04360.04360.0436.

STEP 10

The probability that the proportion of rooms booked in a sample of 480 rooms would be less than 15% is approximately **0.0436**.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? If 18% of hotel rooms should be booked, what's the chance that less than 15% are booked if we check 480 rooms? Watch out! Don't mix up percentages and proportions!
Also, remember we're dealing with a sample of rooms, not the entire hotel.

1. Set up the problem
2. Calculate the standard deviation
3. Calculate the z-score
4. Find the probability

The standard deviation of the sample proportion tells us how spread out our sample results are likely to be.
The formula is: $\sigma_{\hat{p}} = \sqrt{\frac{p \cdot (1 - p)}{n}}$

The z-score tells us how many standard deviations away our target sample proportion ( $0.15$ ) is from the expected population proportion ( $0.18$ ).
The formula is: $z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}$

Plugging in our values: $z = \frac{0.15 - 0.18}{0.0175} = \frac{-0.03}{0.0175} \approx -1.71$ Our z-score is approximately $-1.71$ .

Now, we look up our z-score of $-1.71$ in a standard normal distribution table (or use a calculator).
This tells us the probability of getting a z-score less than $-1.71$ .

The table or calculator gives us a probability of approximately $0.0436$ .

The probability that the proportion of rooms booked in a sample of 480 rooms would be less than 15% is approximately 0.0436.