x is normal random variable; mean is 401.81 and standard deviation is 19.66. find value z such that Pr=.91?

Question

Find a positive value Z such Pr(mu-z(sigma) is less than X less than or equal to mu + z (sigma)) = .91

Answer 1

If you have a table of normal values, look up area .9100 in the body of the table and read off the value of z

(if your table gives areas to the right of z= 0, find the area .4100 in the body of the table.)

Answer 2

You are looking for the values that separate the most extreme 9% from the middle 91%. So, you should draw your normal distribution curve. Shade in the middle 91%. This leaves 9% to be evenly divided into the left and right tail. Thus, the left tail area that is not shaded in is .045 (this is half of .09). Look this AREA up in your standard normal distribution chart to find the corresponding z-score. I find the z-score is about half way between z=-1.69 and z= -1.70. So I split the difference and come up with z = - 1.695. The z you are being asked to find is this z-score. Since it wants to know what positive value z must have, your final answer should be z = 1.695. Hope this helps.

Answer 3

X~N(30, 2^2) non-end distribution, ignore approximately = sign. P(27 < X< 34) = P(z1 < Z< z2), the place z1=(27-30)/2 = -one million.5; z2=(34-30)/2 = 2.0 P(z1 < Z< z2) = P(Z