Question

Continues  probability 

Answer 1

Solution :

Given that ,

mean = $\mu$ = 20

variance = 9

standard deviation = $\sigma$ = $\sqrt$ 9 = 3

X$\sim$ N($\mu$ = 20 , $\sigma$ = 3)

a)

P(X $\leq$ 10) = P((X - $\mu$ ) / $\sigma$ $\leq$ (10 - 20) / 3)

= P(z $\leq$ -3.33)   

Using standard normal table,

=0.0004   

Probability = 0.0004

b)

P(15 $\leq$ X $\leq$ 20) = P((15 - 20)/ 3 $\leq$ (X - $\mu$ ) / $\sigma$ $\leq$ (20 - 20) / 3 )

= P( -1.67 $\leq$ z $\leq$ 0)

= P(z $\leq$ 0) - P(z $\leq$ -1.67)

Using standard normal table,  

= 0.5 - 0.0475

= 0.4525

Probability = 0.4525

c)

Given that ,

mean = $\mu$ = 2

variance = 12

standard deviation = $\sigma$ =$\sqrt$12 = 3.4641

Y$\sim$ N($\mu$ = 2 , $\sigma$ = 3.4641)

Z = 2X + Y + 3

E(Z) = E( 2X + Y + 3)

= E(2X) + E(Y) + 3 by additive property of expectation

=2E(X) + E(Y) + 3

= 2*20 + 2 + 3

= 40+5

= 45

E(Z) = 45

V(Z) = V(2X + Y + 3)

= V(2X) + V(Y) +0 , by additive property of variance.

Since X and Y are independent to each other.

= 4V(X) + 12

=4*9 + 12

=36+12

= 48

V(Z) = 48