Private nonprofit four-year colleges charge, on average, $26,190 per year in tuition and fees. The standard deviation is $6,768. Assume the distribution is normal. Let X be the cost for a randomly selected college. Round all answers to 4 decimal places where possible.

a. What is the distribution of X? X ~ N(
,
)

b. Find the probability that a randomly selected Private nonprofit four-year college will cost less than 28,251 per year. 


c. Find the 72nd percentile for this distribution. $
 (Round to the nearest dollar.)


Given the problem where $ X $ is the cost for a randomly selected private nonprofit four-year college, which follows a normal distribution with a mean $ \mu = 26,190 $ and a standard deviation $ \sigma = 6,768 $, we can answer each part step by step.

### a. What is the distribution of X?
Since $ X $ follows a normal distribution, we write the distribution as:
\[
X \sim N(\mu, \sigma^2)
\]
Here, $ \mu = 26,190 $ and $ \sigma = 6,768 $, so:
\[
X \sim N(26,190, 6,768^2)
\]
The full form of the distribution is:
\[
X \sim N(26,190, 45,809,024)
\]

### b. Find the probability that a randomly selected Private nonprofit four-year college will cost less than $28,251 per year.

To find the probability that $ X $ is less than 28,251, we need to standardize the value using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
For $ X = 28,251 $, $ \mu = 26,190 $, and $ \sigma = 6,768 $:
\[
Z = \frac{28,251 - 26,190}{6,768} = \frac{2,061}{6,768} \approx 0.3044
\]

Next, we look up the Z-score $ Z = 0.3044 $ in the standard normal distribution table or use a calculator to find the probability:
\[
P(X < 28,251) = P(Z < 0.3044) \approx 0.6197
\]
So, the probability is approximately 0.6197.

### c. Find the 72nd percentile for this distribution.

To find the 72nd percentile, we first determine the Z-score corresponding to the 72nd percentile using a standard normal distribution table or a calculator. The Z-score for the 72nd percentile is approximately 0.5842.

Now, we use the Z-score formula to find the corresponding value of $ X $:
\[
X = \mu + Z \cdot \sigma
\]
Substituting the values:
\[
X = 26,190 + 0.5842 \cdot 6,768 \approx 26,190 + 3,956 \approx 30,146
\]

So, the 72nd percentile is approximately **$30,146**.

---

Do you have any questions or would you like more details on any part of the solution?

### Related Questions:
1. What is the probability that a college costs more than $30,000 per year?
2. How much would a college in the 90th percentile cost?
3. What is the interquartile range of the college costs?
4. If a college costs $20,000, what percentile is it in?
5. How would the distribution change if the standard deviation increased by 10%?

**Tip:** Understanding Z-scores can help you calculate probabilities and percentiles in any normal distribution easily.

Private nonprofit four-year colleges charge, on average, $26,190 per year in tuition and fees. The standard deviation is $6,768. Assume the distribution is normal. Let X be the cost for a randomly selected college.

 a. What is the distribution of X? X ~ N( , )

 b. Find the probability that a randomly selected Private nonprofit four-year college will cost less than $28,251 per year.

 c. Find the 72nd percentile for this distribution.

Explore the normal distribution of tuition costs for private nonprofit four-year colleges, with a mean of $26,190 and a standard deviation of $6,768. Learn how to calculate the probability that tuition is less than $28,251 and find the 72nd percentile, using key statistical formulas like Z-scores and percentiles.

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