Question

A random variable is normally distributed with a mean of 30 and a standard deviation of 5. If an observation is randomly selected from the​ distribution, a. What value will be exceeded 15​% of the​ time? b. What value will be exceeded 80​% of the​ time? c. Determine two values of which the smaller has 5​% of the values below it and the larger has 5​% of the values above it. d. What value will 20​% of the observations be​ below?

Answer 1

let xnNCel = 30, 62-52 of we have to bind a value which exceeded art of the time. in. we have to bind 30% such theet pc x - x = 0.15 Plxall x - 30 ) = 0.15 PC 3 x-30) 2015 p(2230) 2085 R-software qnoem Co.85721036 2-30 = 1.036 1. = 35.18 6 we have to find a such that P(x > x = 0.80 P(xa ar 30 ) = 08. .P(22 -30 ) = 0.20 h-software gnoem (0.20 -0842

X-30 – – 0.842 x = 25.79 ③ we have to find tuo valles such that the smallee has 5 l. of the valles below it and the lokgee has 5% of the vallies cebove it. > PC LxLx2 =o.go. P(X24) - 0.05 and 0.06. Plx > 2) = XLx=0.05 - P(x H L - 30 . PC 24 24-30 -0.05 - -0.05 K sootweel, aholem co.05 ) = -1.64 30 = -1.64 C 24 = 2 x x ) .8 - оос) ою, 2 a plx - M L 12-30 )=0.95 PC 2 L 12-30 =0.95 I : R-softucce, a noem co.us) = 1.64. 2-30 = 1.64 .X2 = 38.2 8 xy = 21.8 S T

d Date: 1 Now we have to find value such that 20 7. of the obsecrations be below ie. we have to bind a such that P(XL) = 0.20 P(X-Me X-30) 0.20 :P( 2 L x-30) = 0.20 R-software, gnoem (o. 20) = -0842 z = a - 0.842. 0 = 25.79

a. 35.18

b. 25.79

c. 21.8 & 38.2

d. 25.79