Range of scores making up the interquartile ranger

1 answer:

4 0

For a standard normally distributed random variable Z (with mean 0 and standard deviation 1), we get a probability of 0.0625 for a z-score of Z ≈ 1.53, since

P(Z ≥ 1.53) ≈ 0.9375

You can transform any normally distributed variable Y to Z using the relation

Z = (Y - µ) / σ

where µ and σ are the mean and standard deviation of Y, respectively.

So if s is the standard deviation of X, then

(250 - 234) / s ≈ 1.53

Solve for s :

16/s ≈ 1.53

s ≈ 10.43

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