Question

It follows a regular distribution with an average bond price of 100 and a variance of 9. Which best represents the probability that the bond price is greater than 106 (where z is the probability variable in the standard normal distribution).)
1.p(z<1) 2.p(z>1) 3.p(z>2) 4.p(z<2) 5.p(z=2)

Answer 1

Solution :

P(x > 106) = 1 - P(x < 106)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (106 - 100) / 3]  

= 1 - P(z < 2)

= P(z > 2)

3)