Find the Z-score that separate the middle 63% of the distribution from the area in the tails of the standard normal distribution

Oh, you've stumbled into the world of statistical mischief, my friend! Well, the Z-score that separates the middle 63% of the distribution from the area in the tails of the standard normal distribution is approximately 0.670. But remember, the true entertainment lies in deciphering what that number actually means!

To find the Z-score that separates the middle 63% of the distribution from the area in the tails of the standard normal distribution, we need to find the Z-scores corresponding to the lower and upper percentiles.

Step 1: Determine the area in the tails
Since we want the middle 63% of the distribution, we need to find the area outside of this range. The tails area will be (100% - 63%) / 2 = 18.5% on each side.

Step 2: Find the Z-score for each tail area
Using a Z-table or a calculator, we can find the critical Z-value that corresponds to the area in each tail. For a 0.185 tail area, the Z-score is approximately 0.89.

Step 3: Calculate the Z-score for the middle 63%
The Z-score that separates the middle 63% from the tails can be calculated using the formula:
Z-score = +/- Z-value for tails area

Therefore, the Z-score that separates the middle 63% of the distribution from the area in the tails of the standard normal distribution is approximately +/- 0.89.

To find the Z-score that separates the middle 63% of the distribution from the area in the tails of the standard normal distribution, we can use the properties of the standard normal distribution.

The total area under the standard normal curve is 1, representing 100% of the distribution. Since we want to find the Z-score that separates the middle 63% of the distribution, we know that there will be (100% - 63%) / 2 = 18.5% in each tail of the distribution.

To find the Z-score corresponding to a specific percentile, we can use the cumulative distribution function (CDF) of the standard normal distribution, which gives us the area to the left of a given Z-score.

Using a standard normal table or a statistical calculator, we want to find the Z-score corresponding to the cumulative probability of 0.5 + (0.185 / 2) = 0.5925. This value represents the cumulative probability up to the 59.25th percentile on the normal distribution.

Using a standard normal table or a statistical calculator, we can find that the Z-score corresponding to a cumulative probability of 0.5925 is approximately 0.27.

Therefore, the Z-score that separates the middle 63% of the distribution from the area in the tails of the standard normal distribution is approximately 0.27.

you want z such that P(Z>z) = .63/2

z = .896

So, the scores are ±0.896