i Give the name of the distribution of X if it has a name ii

(i) Give the name of the distribution of X (if it has a name), (ii) find the expected value and variance, and (iii) calculate P( 1 X 2 ) when the moment-generating function of X is given by M(t) = (0.3 + 0.7e^t)⁵

Solution

(i) name of distribution is binomial distribution

(ii) here n=5,p=0.7,q=0.3

expected value = np=5x0.7=3.5=mean

Variance=npq=5x0.7x0.3=1.05,

standrard deviation SD=square root of variannce=

(iii) P( 1 X 2 )= P( -2.44 Z -1.46 )=0.0662

where for X=1, Z=(X-mean)/S.D.=-2.44

for X=2, Z=-1.46

now  P( -2.44 Z -1.46 )=P( -2.44 Z) - P( -1.46 Z)=0.9927-0.9265=0.0662