Suppose that the auto loans from a bank are normally distributed with a mean of $23,334 and a standard deviation of 3,412. what is the probability that a randomly selected loan will be for less than $26,000? answer to three decimal places if necessary.

To illustrate the information in the correct syntax:
X~N(23334, 3412²)
What we're being asked to find is P(X ≤ 26000).
In order to do this, we need to convert this value into one we can look up on the table for the standard normal distribution (Z).
To do this we use the formula:
Z = (X - μ)/σ
Z = (26000 - 23334)/3412 = 0.7813599... ⇒ 0.78
We can say:
P(X ≤ 26000) = P(Z ≤ 0.78) = 0.7823

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raphealnwobi raphealnwobi · Answer 2 · 2022-03-01T12:25:43+01:00

The probability that a randomly selected loan will be for less than $26,000 is 78.23%.

Z score

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (x - μ) / σ

where x is the raw score, μ is the mean and σ is the standard deviation.

Given that μ = 23334, σ = 3412, hence:

for x < 26000:

z = (26000 - 23334) / 3412 = 0.78

P(z < 0.78) = 0.7823 = 78.23%

The probability that a randomly selected loan will be for less than $26,000 is 78.23%.

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