Question #76846

A population has a mean of 220 and a standard deviation of 80. Suppose a sample of size 100 is selected and x ̅ is used to estimate μ.

What is the expected value of x ̅ ?






What is the standard deviation of x ̅?







What is the probability that the sample mean will be within ± 5 of the population mean?








What is the probability that the sample mean will be within ± 10 of the population mean?
1

Expert's answer

2018-05-07T05:46:08-0400

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 xˉ\bar{x} is used to estimate μ\mu.

Question

a. What is the expected value of xˉ\bar{x}?

Solution

E(xˉ)=μ=220E(\bar{x}) = \mu = 220

Question

b. What is the standard deviation of xˉ\bar{x}?

Solution

σ1=σn=80100=8\sigma_1 = \frac{\sigma}{\sqrt{n}} = \frac{80}{\sqrt{100}} = 8

Question

c. What is the probability that the sample mean will be within ±5\pm 5 of the population mean?

Solution

z(215)=21522080/100=0.63z(215) = \frac{215 - 220}{80/\sqrt{100}} = -0.63z(225)=22522080/100=0.63z(225) = \frac{225 - 220}{80/\sqrt{100}} = 0.63p(215<x<225)=p(0.63<z<0.63)=0.73570.2643=0.4714p(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 of the population mean?

Solution

z(210)=21022080/100=1.25z(210) = \frac{210 - 220}{80/\sqrt{100}} = -1.25z(230)=23022080/100=1.25z(230) = \frac{230 - 220}{80/\sqrt{100}} = 1.25p(210<x<230)=p(1.25<z<1.25)=0.89440.1056=0.7888p(210 < x < 230) = p(-1.25 < z < 1.25) = 0.8944 - 0.1056 = 0.7888


Answer provided by https://www.AssignmentExpert.com

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS