In: Statistics and Probability
Find the interval [ μ−z σn√,μ+z σn√μ−z σn,μ+z σn ] within
which 95 percent of the sample means would be expected to fall,
assuming that each sample is from a normal population.
(a) μ = 179, σ = 10, n
= 33. (Round your answers to 2 decimal
places.)
The 95% range is from to .
(b) μ = 867, σ = 18, n
= 10. (Round your answers to 2 decimal
places.)
The 95% range is from to .
(c) μ = 63, σ = 4, n =
27. (Round your answers to 3 decimal
places.)
The 95% range is from to .
Solution :
Given that,
a) mean = = 179
standard deviation = = 10
n = 33
= = 179
= / n = 10 / 33 = 1.74
P( - 2 < < + 2 ) = 95%
= P( 179 - 2 * 1.74 < < 179 + 2 * 1.74 ) = 95%
= P( 179 - 3.48 < < 179 + 3.48 ) = 95%
=P( 175.52 < < 182.48 ) = 95%
b) mean = = 867
standard deviation = = 18
n = 10
= = 867
= / n = 18 / 10 = 5.69
P( - 2 < < + 2 ) = 95%
= P( 867 - 2 * 5.69 < < 867 + 2 * 5.69 ) = 95%
= P( 867 - 11.38 < < 867 + 11.38 ) = 95%
=P( 855.62 < < 878.38 ) = 95%
c) mean = = 63
standard deviation = = 4
n = 27
= = 63
= / n = 4 / 27 = 0.770
P( - 2 < < + 2 ) = 95%
= P( 63 - 2 * 0.770 < < 63 + 2 * 0.770 ) = 95%
= P( 63 - 1.540 < < 63 + 1.540 ) = 95%
=P( 61.460 < < 64.540 ) = 95%