Question

Assume that the age at onset of a certain disease is distributed normally with a mean of 43 years and a variance of 177.69 years.

a) What is the probability that an individual afflicted with the disease developed it before age 31?
probability =  

b) What is the probability that an individual afflicted with the disease developed it after age 48?
probability =  

c) What is the probability that an individual afflicted with the disease developed it between ages 31 and 48?
probability =  

Answer 1

Solution :

Given that ,

mean = = 43

variance = ² = 177.69

standard deviation = = ² = 177.69 = 13.33

a) P(x > 31) = 1 - p( x< 31)

=1- p P[(x - ) / < (31 - 43) / 13.33]

=1- P(z < -0.90)

Using z table,

= 1 - 0.1841

= 0.8159

b) P(x < 48) = P[(x - ) / < (48 - 43) / 13.33 ]

= P(z < 0.38)

Using z table,

= 0.6480

c) P( 31 < x < 48) = P[(31 - 43) / 13.33) < (x - ) /  < (48 - 43) / 13.33) ]

= P( -0.90 < z < 0.38)

= P(z < 0.38 ) - P(z < -0.90)

Using z table,

= 0.6480 - 0.1841

= 0.4639