Question

Worldwide life expectancy for women (5-year average between 2010-2015): mean=73.25, SD=7.65.

1. What percent of all women will live beyond 60? (.5 pt)
2. What percent of all women will live less than75 years? (.5 pt)
3. What percent of all women will live less than50 years? (.5 pt)
4. What percent of all women will live between 55 to 80 years of age? (.5 pt)

Answer 1

Solution :

Given that ,

mean = = 73.25

standard deviation = = 7.65

a) P(x > 60) = 1 - p( x< 60)

=1- p P[(x - ) / < (60 - 73.25) / 7.65 ]

=1- P(z < -1.73)

Using z table,

= 1 - 0.0418

= 0.9582

The percentage is = 95.82%

b) P(x < 75) = P[(x - ) / < (75 - 73.25) / 7.65]

= P(z < 0.23)

Using z table,

= 0.5910

The percentage is = 59.10%

c) P(x < 50) = P[(x - ) / < (50 - 73.25) / 7.65]

= P(z < -3.04)

Using z table,

= 0.0012

The percentage is = 0.12%

d) P(55 < x < 80) = P[(55 - 73.25)/ 7.65) < (x - ) /  < (80 - 73.25) / 7.65) ]

= P(-2.39 < z < 0.88)

= P(z < 0.88) - P(z < -2.39)

Using z table,

= 0.8106 - 0.0084

= 0.8022

The percentage is = 80.22%