Answers to review problems:
1. p(3) = 0.3
the Sum must equal 1
2. ∑x∙p(x) = (-4)(20/30) + 0(5/30) + 5(4/30) +15(1/30) = -$1.50
VAR(X) = ∑x2p(x- (E(X))2 = (16)(20/30) + 0(5/30) + 25(4/30) + 225 (1/30) - ($1.50)2 =
= 21.5 - 2.25 = 19.25
Standard deviation = $4.39
3. X= # of pet owners
p= 0.25, q = 0.75, n = 18
P(x=5) = binompdf(18,.25,5) = 0.1988
P(x<10) = P(x≤9) = binomcdf(18,0.25,9) = 0.9946
P(x≥10) =1 - P(x≤9) = 1-0.9946 =0.0054
Mean = np = 18*0.25 = 4.5
VARIANCE = npq = 18*0.25*0.75 = 3.375 so the Standard deviation is 1.84
4. P(z>-2.40) =normalcdf(-2.4, big#,0,1)= 0.9918
5. P(-1<z<2.3) = normalcdf(-1, 2.3,0,1)= 0.8306
P(z<1.96) = normalcdf(small#, 1.96,0,1)= 0.975
6. P(25<x< 38) Two ways: convert to "Z" by (x-mean)/(stand. dev)
P(-.833<z<1.33)
normalcdf( -.83,1.33) = .7064
normalcdf(25,38,30,6,) = 0.7064
7. P(X>30) normalcdf(30, big#, 24.5,5.5) = 0.1587
8. InvNorm(0.20) = -0.84
InvNorm(0.95) = 1.645 use z = 1.645 = (x-80)/8 and solve for x or
InvNorm(0.95, 80, 8) = 93.16
9. a. 3 standard deviations≈ 99.7%
b. 2 standard deviations≈95%
c. 1 standard deviation≈68%
[From calculator normalcdf: .... 0.9973,
0.9545, 0. 68270]