Answer on Question #76846 – Math – Statistics and Probability
A population has a mean of 220 and a standard deviation of 80. Suppose a sample of size 100 is selected and x ˉ \bar{x} x ˉ is used to estimate μ \mu μ .
Question
a. What is the expected value of x ˉ \bar{x} x ˉ ?
Solution
E ( x ˉ ) = μ = 220 E(\bar{x}) = \mu = 220 E ( x ˉ ) = μ = 220 Question
b. What is the standard deviation of x ˉ \bar{x} x ˉ ?
Solution
σ 1 = σ n = 80 100 = 8 \sigma_1 = \frac{\sigma}{\sqrt{n}} = \frac{80}{\sqrt{100}} = 8 σ 1 = n σ = 100 80 = 8 Question
c. What is the probability that the sample mean will be within ± 5 \pm 5 ± 5 of the population mean?
Solution
z ( 215 ) = 215 − 220 80 / 100 = − 0.63 z(215) = \frac{215 - 220}{80/\sqrt{100}} = -0.63 z ( 215 ) = 80/ 100 215 − 220 = − 0.63 z ( 225 ) = 225 − 220 80 / 100 = 0.63 z(225) = \frac{225 - 220}{80/\sqrt{100}} = 0.63 z ( 225 ) = 80/ 100 225 − 220 = 0.63 p ( 215 < x < 225 ) = p ( − 0.63 < z < 0.63 ) = 0.7357 − 0.2643 = 0.4714 p(215 < x < 225) = p(-0.63 < z < 0.63) = 0.7357 - 0.2643 = 0.4714 p ( 215 < x < 225 ) = p ( − 0.63 < z < 0.63 ) = 0.7357 − 0.2643 = 0.4714 Question
d. What is the probability that the sample mean will be within ± 10 \pm 10 ± 10 of the population mean?
Solution
z ( 210 ) = 210 − 220 80 / 100 = − 1.25 z(210) = \frac{210 - 220}{80/\sqrt{100}} = -1.25 z ( 210 ) = 80/ 100 210 − 220 = − 1.25 z ( 230 ) = 230 − 220 80 / 100 = 1.25 z(230) = \frac{230 - 220}{80/\sqrt{100}} = 1.25 z ( 230 ) = 80/ 100 230 − 220 = 1.25 p ( 210 < x < 230 ) = p ( − 1.25 < z < 1.25 ) = 0.8944 − 0.1056 = 0.7888 p(210 < x < 230) = p(-1.25 < z < 1.25) = 0.8944 - 0.1056 = 0.7888 p ( 210 < x < 230 ) = p ( − 1.25 < z < 1.25 ) = 0.8944 − 0.1056 = 0.7888
Answer provided by https://www.AssignmentExpert.com
Comments