Question

Given a random variable XX following normal distribution with mean of -3 and standard deviation of 4. Then random variable Y=0.4X+5Y=0.4X+5 is also normal.

(1)(2pts) Find the distribution of YY, i.e. μY,σY.μY,σY.

(2)(3pts) Find the probabilities P(−4<X<0),P(−1<Y<0).P(−4<X<0),P(−1<Y<0).

(3)(3pts) Find the probabilities P(−4<X¯<0),P(3<Y¯<4).P(−4<X¯<0),P(3<Y¯<4).

(4)(4pts) Find the 53th percentile of the distribution of X.

Answer 1

Given a random variable X following normal distribution with mean of -3 and standard deviation of 4. Then random variable Y=0.4X+5 is also normal.

(1) The distribution of Y is computed as follows:

M(Y) = Mean of Y = Mean of(0.4X+5) = 0.4Mean(X)+5 = 0.4(-3)+5 =3.8

S(Y)= SD of Y = SD of(0.4X+5) = 0.4SD(X) = 0.4(4)=1.6

(2) Z(x) = (x-mean)/sd = (x-(-3))/4 =(x+3)/4

Z(y) = (y-mean)/sd = (y-3.8)/1.6

To find P(−4<X<0),P(−1<Y<0).

P(−4<X<0) = P((-4+3)/4<Z(X)<(0+3)/4)=P(-0.25<Z(X)< 0.75)=P(Z(X)<0.75)-P(Z(X)<-0.2 5)

= 0.7734-0.4013 = 0.3721

P(−1<Y<0)= P((-1-3.8)/1.6 <Z(Y)<(0-3.8)/1.6) = P(-3 <Z(Y)<-2.375)

=P(Z(Y)<-2.375)-P(Z(Y)<-3) = 0.0088-0.0013 = 0.0075

(3) To find the probabilities P(−4<X¯<0),P(3<Y¯<4)

Since, here sample size n is not given then it is assumed as n=1

Z(X¯) = (X¯-mean)sqrt(n)/sd = (X¯-(-3))sqrt(1)/4 =( X¯+3)/4

Z(Y¯) = (Y¯-mean) sqrt(n)/sd = (Y¯-3.8) sqrt(1)/1.6 = (Y¯-3.8)/1.6

P(−4<X¯<0) = P(( -4+3)/4<Z(X¯)<(0+3)/4)= P(-0.25<Z(X¯)< 0.75)

=P(Z(X¯)<0.75)-P(Z(X¯)<-0.2 5) = 0.7734-0.4013 = 0.3721

P(3<Y¯<4)= P((3-3.8)/1.6<Z(Y¯)<(4-3.8)/1.6)) = P(-0.5<Z(Y¯)<0.125)

= P(Z(Y¯)<0.125)-P(Z(Y¯)<-0.5) = 0.54975-0.3085 = 0.24125

(4) 53rd percentile p53of the distribution of X is found such that

P(X<p53)=53% = 0.53

P(Z(X)< (p53+3)/4) = 0.53

P(Z(X)< 0.075) = 0.53(approximately from the area tables of standard normal distribution)

Therefore, (p53+3)/4 =0.075

p53+3= 0.075x4 = 0.3

p53 = 0.3-3=-2.7

Hence, the 53rd percentile of the distribution of X is -2.7