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