a survey of 20 colleges found that seniors graduated with an average $12000 in debt from student loans. The debt was normally distributed with a standard deviation of $3200. Find the probability that a senior graduated owing more than $16000.
16000 is 4000 above the average
4000/3200 = 1.2 sd
use a z-score table to find the probability
Well, if we assume that student loan debt is a funny thing (which it isn't, trust me), let's dive into some jocularity while tackling this probability problem.
We know that the average student loan debt is $12,000, with a standard deviation of $3,200. So, if we were to organize a comedy roast for these seniors, with their debts representing the jokes, we'd need to find out the probability that a senior owes more than $16,000.
To calculate this probability, we need to utilize the normal distribution and its lovely z-scores. Here's the formula for finding the z-score:
z = (x - μ) / σ
where x is the value we're interested in (in this case, $16,000), μ is the mean ($12,000), and σ is the standard deviation ($3,200).
Plugging in the numbers, we get:
z = (16,000 - 12,000) / 3,200
Simplifying, we find:
z ≈ 1.25
Now, we need to find the probability associated with this z-score. Using a z-table or a nifty online calculator, we can determine that the probability of a senior graduating with debt over $16,000 is about 0.1056, or approximately 10.56%.
So, it seems that 10.56% of seniors have debts so high, they could be the punchline of an unfortunate finance-themed joke. Remember, folks, student loan debt is no laughing matter, but sometimes a little humor can help us make sense of statistics.
To find the probability that a senior graduated owing more than $16000, we need to calculate the area under the normal distribution curve to the right of $16000.
Let's denote the average debt as μ = $12000, and the standard deviation as σ = $3200.
To proceed, we will use the z-score formula:
z = (x - μ) / σ,
where x is the value we want to find the probability for.
Substituting the given values:
z = (16000 - 12000) / 3200,
z = 4000 / 3200,
z = 1.25.
Now, we need to find the area to the right of this z-score using a Z-table or a calculator.
Looking up the z-score of 1.25 in the Z-table, we find that the corresponding area is 0.8944.
However, we're interested in the probability of owing more than $16000, so we subtract this value from 1:
P(X > 16000) = 1 - 0.8944,
P(X > 16000) ≈ 0.1056.
Therefore, the probability that a senior graduated owing more than $16000 is approximately 0.1056 or 10.56%.
To find the probability that a senior graduated owing more than $16000, we need to use the normal distribution and calculate the z-score.
The z-score tells us how many standard deviations away from the average (mean) a particular value is.
First, we need to calculate the z-score for $16000 using the formula:
z = (X - μ) / σ
where:
X = value we are interested in (in this case, $16000)
μ = mean value (in this case, $12000)
σ = standard deviation (in this case, $3200)
Substituting the given values:
z = (16000 - 12000) / 3200
z = 4000 / 3200
z = 1.25
Next, we need to find the probability corresponding to this z-score using the Z-table (also known as the standard normal table). This table provides the area under the normal distribution curve for different z-scores.
However, since we are interested in finding the probability of a value greater than $16000, we need to find the area to the right of the z-score. We can do this by subtracting the cumulative probability (area to the left) from 1.
Using the Z-table, the cumulative probability corresponding to a z-score of 1.25 is approximately 0.8944. Therefore, the probability of owing more than $16000 is:
P(X > 16000) = 1 - P(X ≤ 16000)
P(X > 16000) = 1 - 0.8944
P(X > 16000) ≈ 0.1056
Thus, the probability that a senior graduated owing more than $16000 is approximately 0.1056 or 10.56%.