7) 164.26
8)77.08
9)2405.364
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steps for 8)
The water level at a local pier rises and falls with the tide. Yesterday, the maximum depth of the water at the pier was 8 feet,
and the minimum depth was 4 feet. High tide occurred at 12:00 am and low tide occurred at 12:20 pm. Which function models the depth, in feet, of the water at the pier yesterday, as a function of the time t in minutes past high tide?
1 answer:
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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:
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)= 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!