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Morgarella [4.7K]
3 years ago
8

The answer to it now

Mathematics
1 answer:
kondaur [170]3 years ago
8 0
I think it may be B.
 (not sure)
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Answer:

D. The center remains constant, and the area in the tails of the distribution decreases.

Step-by-step explanation:

Hello!

Be it two independent random variables, X~N(μ;σ²) and U~Xₙ², the variable t is determined by the quotient between a random variable N(0;1) and the square root of a Chi-Square variable divided by its degrees of freedom:

t= \frac{(X-Mu)/Sigma}{\sqrt{U/n} }

As a consequence of this, the structure of the distribution depends on the parameter n (degrees of freedom), it is centered in zero and has a bell-shape similar to the normal distribution.

It has a mean and variance:

E(t)= 0 for n > 1

V(t)= \frac{n}{n - 2} n > 2

As you can see the variance of the distribution is directly affected by its degrees of freedom, which means that when the degrees of freedom change, the variance of the distribution change and so does its shape.

When ↑n ⇒ ↑V(t) ⇒ The area under the tails increases.

When ↓n ⇒ ↓V(t) ⇒ The area under the tails decreases.

In this example, the degrees of freedom of the distribution decreased from 40 to 20, then the variance of the distribution decreases and it "flattens", i.e. the area under the tails gets lowered.

The E(t) isn't affected by the modification of n, so the distribution remains centered in zero.

I hope this helps!

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