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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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The Angle HCA is equal to 52°. This is arrived at using the knowledge of the Total value of Angles in a Triangle and the Total Value of Angles in a Polygon.

<h3>What other principles were used to arrive at the answer?</h3>

The other principles of mathematics that were used to arrive at the above answer are:

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Step 1 - Recall that we have been given Angles AHB and BAH to be 128° and 28° respectively.

We also know that:

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Since A...therefore:

When ∠AHB (128°) and ∠BAH are taken from 180° we have DBA = ∠28°.

By observation, we can deduce that ∠BDE, ∠CDH, ∠CEH and ∠AEH are all right-angled triangles.


Using the above, we are able to repeat this process of solving for each angle until we have ∠HCA.

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That is, 90+ 128 + 90 + 52 = 360°

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