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)
![W = \frac{X-\mu}{\sigma} = \frac{X-70}{4}](https://tex.z-dn.net/?f=%20W%20%3D%20%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%20%3D%20%5Cfrac%7BX-70%7D%7B4%7D%20)
The values of the cumulative distribution function of W, denoted by
, can be found in the attached file. Lets call
. We have
![0.1 = P(X > z) = P(\frac{X-70}{4} > \frac{z-70}{4}) = P(W > y) = 1-\phi(y)](https://tex.z-dn.net/?f=0.1%20%3D%20P%28X%20%3E%20z%29%20%3D%20P%28%5Cfrac%7BX-70%7D%7B4%7D%20%3E%20%5Cfrac%7Bz-70%7D%7B4%7D%29%20%3D%20P%28W%20%3E%20y%29%20%3D%201-%5Cphi%28y%29)
Thus
![\phi(y) = 1-0.1 = 0.9](https://tex.z-dn.net/?f=%20%5Cphi%28y%29%20%3D%201-0.1%20%3D%200.9%20)
by looking at the table, we find that y = 1.28, therefore
![\frac{z-70}{4} = 1.28\\z = 1.28*4+70 = 75.12](https://tex.z-dn.net/?f=%5Cfrac%7Bz-70%7D%7B4%7D%20%3D%201.28%5C%5Cz%20%3D%201.28%2A4%2B70%20%3D%2075.12)
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] .