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I have a question regarding how to answer this: Q: A credit risk study found that an individual with good credit score has an average debt of 15,000. If the debt of an individual with good credit score is normally distributed with standard deviation $3000, determine the shortest interval that contains 95% of the debt values.

Lower bound___ (enter value rounded to the nearest dollar; do not enter the $ sign)

Upper bound___ (enter value rounded to the nearest dollar; do not enter the $ sign)

I have tried to calculate the lower limit as: -1.645 * (3000) + 15000 and the upper limit as: 1.645 * (3000) + 15000, however, the answers which I have derived for this question appear to be incorrect. But I have no clue as to where the problem arises from.

Hizumaru
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1 Answers1

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It turns out you used the z-score that corresponds to an area of 90%. Let me explain.

The desired area is $0.95$ which means $5\%$ of the data should not be contained in the interval. Because the normal distribution exhibits symmetry, you can say that half of that five percent is in each tail. So, we need to find the z-score that corresponds to an area of $0.025$, which is approximately $-1.96$.

So the desired interval is $15000 \pm (1.96 \cdot 3000) = (9120, 20880).$

Sean Roberson
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  • Hi! I have tried this answer, however, it still says that it is wrong. – Hizumaru May 21 '23 at 00:20
  • This sounds like a homework question in an online system. Edit your original post to include the directions, word by word, including any rounding rules. – Sean Roberson May 21 '23 at 00:23
  • For both the lower and upper limit: enter value rounded to the nearest dollar; do not enter the $ sign) It also italicized "shortest" interval within the question – Hizumaru May 21 '23 at 00:26
  • Aight never mind it's correct. Thank you! – Hizumaru May 21 '23 at 01:56