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Crank
2 years ago
6

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

SAT
1 answer:
Vikki [24]2 years ago
8 0

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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One of the examples from the text that can be regarded as clear, logical and sincere is:

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The president used the words such as revive to show that the nation was still capable of enduring and coming out of the dark times that they were in.

From these words we can see that the president was logical and also sincere in what they were passing through and how they were to overcome it.

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These are words that are said to convey the truth about given situations in simple and well defined terms.

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Read more on President Roosevelt's speech:

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First we need to convert it into normal distribution.

From the attached file, we can see the shaded area we are looking for.

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P(X > 15884) = P(Z > 0.59).

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