D^-2(d^3)^4/d^5= ??? I think it is d^5 but I’m not sure
1 answer:
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Answer:
The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12]
Step-by-step explanation:
I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.
In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1
If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)
The values of the cumulative distribution function of W, denoted by , can be found in the attached file. Lets call
. We have
Thus
by looking at the table, we find that y = 1.28, therefore
The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.
The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12] .