Question

The life expectancy in the United States is 75 with a standard deviation of 7 years. A random sample of 49 individuals is selected. Round all probabilities to four decimal places.

What is the probability that the sample mean will be larger than 77 years? Answer

What is the probability that the sample mean will be within 1 year of the population mean? Answer

What is the probability that the sample mean will be within 2.5 years of the population mean? Answer

Answer 1

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

Mean = 75 years

Samole size, n = 49

Standard error =

=

= 1

P(sample mean will be larger than 77 years) = P( > 77)

= 1 - P( < 77)

= 1 - P(Z < (77 - 75)/1)

= 1 - P(Z < 2)

= 1 - 0.9772

= 0.0228

P(sample mean will be within 1 year of the population mean) = P(74 < < 76)

= P( < 76) - P( < 74)

= P(Z < (76 - 75)/1) - P(Z < (74 - 75)/1)

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

= 0.8413 - 0.1587

= 0.6826

P(sample mean will be within 2.5 years of the population mean) = P(72.5 < < 77.5)

= P( < 77.5) - P( < 72.5)

= P(Z < (77.5 - 75)/1) - P(Z < (72.5 - 75)/1)

= P(Z < 2.5) - P( < -2.5)

= 0.9938 - 0.0062

= 0.9876