The borrowers with good credit scores the mean debt for savings account is k15015. Assume the standard deviation is k3540 and that debt amounts are normally distributed.
What is the probability that the debt for a borrower with good credit is more than k8000?
(p - Value for Z = 0.84322 is 0.2995)
To solve this problem, we need to calculate the standard score or z-score for the debt amount of k8000.
The z-score formula is given by:
z = (x - μ) / σ
where x is the given debt amount, μ is the mean debt, and σ is the standard deviation.
In this case, x = k8000, μ = k15015, and σ = k3540.
Plugging in these values, we have:
z = (8000 - 15015) / 3540
z ≈ -1.3746
Now we need to find the probability of a z-score being greater than -1.3746, which is equivalent to finding 1 - P(z < -1.3746).
Using the p-value for a z-score of 0.84322, which is 0.2995, we can deduce that the probability of a z-score being less than -0.84322 is 0.5 - 0.2995 = 0.2005.
Thus, P(z < -1.3746) ≈ 0.2005.
Finally, the probability that the debt for a borrower with good credit is more than k8000 is given by:
P(debt > k8000) = 1 - P(z < -1.3746)
P(debt > k8000) ≈ 1 - 0.2005
P(debt > k8000) ≈ 0.7995