Question
Find the zscores that bound the middleb94% of the area under a standard curve. Enter in ascending order. Round to 2 decimal places
Answers
GPT-4o mini
To find the z-scores that bound the middle 94% of the area under a standard normal curve, you first need to determine the tails of the distribution. Since 94% of the area is in the middle, the remaining area in the tails is:
\[
100\% - 94\% = 6\%
\]
Since this area is distributed equally in the two tails, each tail will have:
\[
\frac{6\%}{2} = 3\%
\]
Now, you need to find the z-scores that correspond to the cumulative area of 3% in the left tail and 97% in the right tail (i.e., \(100\% - 3\%\)).
Using a standard normal distribution table or a calculator:
1. **For the lower z-score (3% cumulative):** This corresponds to approximately \(z = -1.88\).
2. **For the upper z-score (97% cumulative):** This corresponds to approximately \(z = 1.88\).
Therefore, the z-scores that bound the middle 94% of the area under the standard normal curve in ascending order are:
\[
\boxed{-1.88 \text{ and } 1.88}
\]
\[
100\% - 94\% = 6\%
\]
Since this area is distributed equally in the two tails, each tail will have:
\[
\frac{6\%}{2} = 3\%
\]
Now, you need to find the z-scores that correspond to the cumulative area of 3% in the left tail and 97% in the right tail (i.e., \(100\% - 3\%\)).
Using a standard normal distribution table or a calculator:
1. **For the lower z-score (3% cumulative):** This corresponds to approximately \(z = -1.88\).
2. **For the upper z-score (97% cumulative):** This corresponds to approximately \(z = 1.88\).
Therefore, the z-scores that bound the middle 94% of the area under the standard normal curve in ascending order are:
\[
\boxed{-1.88 \text{ and } 1.88}
\]