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what is the value of 67th percentile in a standard normal distribution?

Sagot :

LammettHashidn

Let Z be a random variable following the standard normal distribution with mean µ = 0 and standard deviation σ = 1.

The 67th percentile of the distribution is the value z that separates the bottom 67% of the distribution from the top 100% - 67% = 33%. In terms of probability, we have

Pr[Z ≤ z] = 0.67

Use the inverse CDF for the normal distribution (or lookup z scores in a table) to find

z = ɸ⁻¹(0.67) ≈ 0.4399

where ɸ(z) is the CDF for the normal distribution.

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