A survey of 20 colleges found that seniors graduated with an average $28,950 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 $15,000.
We can use the standard normal distribution to solve this problem by transforming the original data into z-scores:
z = (x - μ) / σ
where x is the value we are interested in (in this case, $15,000), μ is the mean debt ($28,950), and σ is the standard deviation ($3200).
z = (15000 - 28950) / 3200 = -4.078125
Note that the negative z-score indicates that $15,000 is more than 4 standard deviations below the mean.
Using a standard normal distribution table or calculator, we can find that the probability of a z-score less than -4.08 is very close to 0 (less than 0.0001). Therefore, the probability of a senior graduating with debt exceeding $15,000 is very close to 1 (or 100%). This suggests that the original data may have outliers or other unusual characteristics that make the distribution highly skewed or non-normal.
Well, let's do some math and put on our thinking caps!
Since the debt is normally distributed, we can use the z-score formula to find the probability. The z-score tells us how many standard deviations a data point is away from the mean. In this case, our mean is $28,950 and our standard deviation is $3200.
To find the z-score for $15,000, we can use the formula:
z = (x - μ) / σ
where x is the value we're looking for, μ is the mean, and σ is the standard deviation. Plugging in the numbers, we get:
z = (15,000 - 28,950) / 3200
Calculating this, we find that the z-score is approximately -4.36.
Now, we need to find the probability of a z-score less than -4.36. However, since we want to find the probability of owing MORE than $15,000, we need to calculate the probability of a z-score GREATER than -4.36.
But wait, negative z-scores represent values less than the mean. Since we want values greater than $15,000, we need to find the probability of a z-score LESS than -4.36 and then subtract it from 1 to find the probability of a z-score GREATER than -4.36.
So, now we need to consult a z-table (or calculator) to find the probability associated with a z-score of -4.36. Looking it up, we find that the probability is approximately 0.000000013.
However, since we want the probability of a z-score GREATER than -4.36, we subtract this value from 1:
1 - 0.000000013 = 0.999999987
So, the probability that a senior graduated owing more than $15,000 is approximately 0.999999987, which is almost as rare as finding a unicorn juggling pickle jars on a unicycle. It's extremely unlikely!
To find the probability that a senior graduated owing more than $15,000, we need to calculate the z-score and then find the corresponding probability using the standard normal distribution.
Step 1: Calculate the z-score:
Z = (X - μ) / σ
where X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case, X = $15,000, μ = $28,950, and σ = $3,200.
Z = (15000 - 28950) / 3200
Step 2: Calculate the probability using the z-table or a calculator.
Using a z-table or a calculator, we find that the z-score corresponding to $15,000 is approximately -3.078.
Step 3: Find the probability:
Since we want to find the probability that a senior graduated owing more than $15,000, we need to find the area to the right of the z-score on the standard normal distribution curve.
P(X > $15,000) = 1 - P(Z < -3.078)
Using a z-table or a calculator, we find that P(Z < -3.078) is approximately 0.0011.
Therefore, P(X > $15,000) = 1 - 0.0011 ≈ 0.9989.
Therefore, the probability that a senior graduated owing more than $15,000 is approximately 0.9989, or 99.89%.
To find the probability that a senior graduated owing more than $15,000, we need to calculate the z-score and then use the z-table or calculator to find the corresponding probability.
First, let's calculate the z-score using the formula: z = (x - μ) / σ
Where:
x = the value we want to find the probability for, in this case, $15,000
μ = the mean of the distribution, which is $28,950
σ = the standard deviation, which is $3,200
Now, substitute the values into the formula:
z = (15000 - 28950) / 3200
Simplifying this equation:
z = -13950 / 3200
Calculating this expression:
z ≈ -4.359
Now, we need to find the probability associated with this z-score. We will use a standard normal distribution table or a calculator to find the corresponding probability.
Using a standard normal distribution table, find the area to the left of z = -4.359 (since we are interested in values greater than $15,000).
The table gives us the probability up to a certain z-score. Since we want the probability above -4.359, we need to subtract the cumulative probability from 1.
Using the table or calculator, the probability is approximately 0.99999. Subtracting this from 1, we get:
1 - 0.99999 = 0.00001
So, the probability that a senior graduated owing more than $15,000 is approximately 0.00001, or 0.001%