Question

The life expectancy in the United States is 75 with a standard deviation of 7 years.

a.	For a sample of 49, what is the probability that the sample mean will be within 1.4 years of the mean?
b.	For a sample of 49, what is the probability that the sample mean will be within 2 years of the mean?
c.	For a sample of 81, what is the probability that the sample mean will be within 1.4 years of the mean?
d.	For a sample of 81, what is the probability that the sample mean will be within 2 years of the mean?

Answer 1

= 75 years

= 7 years

For sampling distribution of mean, P( < A) = P(Z < (A - )/)

a) Sample size, n = 49

= = 75 years

=

=

= 1

P(sample mean will be within 1.4 years of the mean) = P(-1.4/ < Z < 1.4/)

= P(Z < 1.4/) - P(Z < -1.4/)

= P(Z < 1.4/1) - P(Z < -1.4/1)

= P(Z < 1.4) - P(Z < -1.4)

= 0.9192 - 0.0808

= 0.8384

b) Sample size, n = 49

= = 75 years

=

=

= 1

P(sample mean will be within 1.4 years of the mean) = P(-2/ < Z < 2/)

= P(Z < 2/) - P(Z < -2/)

= P(Z < 2/1) - P(Z < -2/1)

= P(Z < 2) - P(Z < -2)

= 0.9772 - 0.0228

= 0.9544

c) Sample size, n = 81

= = 75 years

=

=

= 0.7778

P(sample mean will be within 1.4 years of the mean) = P(-1.4/ < Z < 1.4/)

= P(Z < 1.4/) - P(Z < -1.4/)

= P(Z < 1.4/0.7778) - P(Z < -1.4/0.7778)

= P(Z < 1.8) - P(Z < -1.8)

= 0.9641 - 0.0359

= 0.9282

d) Sample size, n = 81

= = 75 years

=

=

= 0.7778

P(sample mean will be within 1.4 years of the mean) = P(-2/ < Z < 2/)

= P(Z < 2/) - P(Z < -2/)

= P(Z < 2/0.7778) - P(Z < -2/0.7778)

= P(Z < 2.57) - P(Z < -2.57)

= 0.9949 - 0.0051

= 0.9898