Question

According to actuarial tables, life spans in the United States are approximately normally distributed with a mean of about 75 years and a standard deviation of about 16 years.

1) Find the probability that a randomly selected person lives between 60 less than 90 years.

2) Find the probability that a randomly selected person lives less than 50 years or more than 100 years.

3) Find the probability that a randomly selected person lives exactly 75 years.

4) What age is considered to be the 99th percentile?

5) What age is considered to be in the 10th percentile?

Answer 1

Solution:
Given:  life spans in the United States are approximately normally distributed with a mean of about 75 years and a standard deviation of about 16 years.

Thus and

Part 1) Find the probability that a randomly selected person lives between 60 less than 90 years.

P( 60 < X < 90) = .....?

Find z scores:

Thus we get:

P( 60 < X < 90) = P( -0.94 < Z< 0.94)

P( 60 < X < 90) = P( Z< 0.94) - P(Z < -0.94 )

Look in z table for z = 0.9 and 0.04 as well as for z = -0.9 and 0.04 and find corresponding area.

From z table , we get:

P( Z< 0.94)=0.8264

and

P( Z< -0.94) = 0.1736

Thus

P( 60 < X < 90) = P( Z< 0.94) - P(Z < -0.94 )

P( 60 < X < 90) = 0.8264 - 0.1736

P( 60 < X < 90) = 0.6528

Part 2) Find the probability that a randomly selected person lives less than 50 years or more than 100 years.

P( X < 50 or X > 100) = ...........?

P( X < 50 or X > 100) = P( X < 50) + P( X > 100)

Find z scores:

Thus we get:
P( X < 50 or X > 100) = P( X < 50) + P( X > 100)

P( X < 50 or X > 100) = P( Z < -1.56) + P( Z > 1.56)

P( X < 50 or X > 100) = P( Z < -1.56) + [ 1 - P( Z < 1.56) ]

Look in z table for z = 1.5 and 0.06 as well as for z = -1.5 and 0.06 and find corresponding area.

From z table , we get:

P( Z < 1.56) = 0.9406 and

P( Z < -1.56) =0.0594

thus

P( X < 50 or X > 100) = P( Z < -1.56) + [ 1 - P( Z < 1.56) ]

P( X < 50 or X > 100) =0.0594 + [ 1 - 0.9406 ]

P( X < 50 or X > 100) =0.0594 + 0.0594

P( X < 50 or X > 100) =0.1188

Part 3) Find the probability that a randomly selected person lives exactly 75 years.

For continuous distribution , we assume there are infinite possible values in sample space, thus probability of any particular value or number is approximately 0.

thus P( X = 75) =0.0000

Part 4) What age is considered to be the 99th percentile?

That is find x value such that:

P(X < x) = 99%

P(X < x) = 0.9900

thus find z value such that:

P( Z < z ) = 0.9900

Thus look in z table for Area = 0.9900 or its closest area and find corresponding z critical value.

Area 0.9901 is closest to 0.9900 , thus corresponding z value is 2.3 and 0.03

Thus z  = 2.33

Now use following formula to find x value:

thus the 99th percentile age is 112.28 years.

Part 5) What age is considered to be in the 10th percentile?

That is find x value such that:

P(X < x) =10%

P(X < x) = 0.1000

thus find z value such that:

P( Z < z ) = 0.1000

Thus look in z table for Area = 0.1000 or its closest area and find corresponding z critical value.

Area 0.1003 is closest to 0.1000 and it corresponds to -1.2 and 0.08

Thus z = -1.28

Now use following formula to find x value:

thus the 10th percentile age is 54.52 years.