Show AH Work!!! 1. Find the probability value: (15 points = 5 points each) a) P(-0.32 5 z 51.62)= b) P(z>-1.03)= e) P(zg-1.35)= 2. Find the value of 20. such that: a) P( -z:: E z 5 20) = 0.82 (5 points) b) PG: E 20) = 0.9082 (5 points) 3. The random variable X is normally distributed with mean 80 and standard deviation 12. a) "What is the probability that a 1Jalue of X chosen at random will be between 7'0 and 85'? (10 points) b) What is the probability that a value of X chosen at random will be less than 75? (IllI points) Formulas Discrete Probability Distribution Mean E(X ) = u = >xP(x) all x Variance v(X)=o' = >(x -u)'P(x) Standard Deviation 0 = VO2 all x Binomial Distribution If X ~ Bin(n, p), then P(X = x) -(20*(1-pax-. n! x!(n-x)[ P (1-p)"-* x=0,1,...,n. Computing Standard Scores Individual Z = - data value-mean standard deviation Group Z = data value-mean standard deviation/ Vn Hypothesis Testing Mean Z = x - H Proportion Z =- P-P pq V InTable of Standard Normal Probabilities for Negative Z-scores Table of Standard Normal Probabilities for Positive Z-scores Note that the probabilities given in this table represent the area to the LEFT of the z-score. The area to the RIGHT of a z-score = 1 - the area to the LEFT of the z-score