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given a normal distribution with a mean 25, what is the standard deviation if 18% of the values are above 29

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Answer:

The standard deviation (σ) of the normal distribution is approximately 4.5455 when 18% of the values are above 29 with a mean of 25.

Explanation:

To find the standard deviation of a normal distribution when you know the mean and the percentage of values above a certain threshold, you can use the Z-score formula and the standard normal distribution table.

First, find the Z-score that corresponds to the 82% percentile (since 100% - 18% = 82% are below 29). You can use a standard normal distribution table or calculator for this.

In this case, you are looking for the Z-score such that P(Z > z) = 0.82. Using a standard normal distribution table or calculator, you can find that the Z-score is approximately 0.88.

Next, use the Z-score formula to relate the Z-score, the mean (μ), and the standard deviation (σ):

Z = (X - μ) / σ

Where:

Z is the Z-score (0.88 in this case)

X is the value (29 in this case)

μ is the mean (25 in this case)

σ is the standard deviation (which we want to find)

Plug in the values into the formula:

0.88 = (29 - 25) / σ

Solve for σ:

0.88 = 4 / σ

To isolate σ, take the reciprocal of both sides:

σ = 4 / 0.88 ≈ 4.5455

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