Question 346588: Can you please tell me if my answers are correct:
Standard deviation of time women with one job are employed during first 8 years of their career is 92 weeks. Length of time employed during the first 8 years of career is a left skewed variable. For that variable:
a. Determine the sampling distribution of the sample mean for simple random samples of 50 women with one job. my answer is 92/√50=13.01 – sampling distribution of the sample mean will be approximately normal because the sample is large enough.
b. Obtain the probability that the sampling error made in estimating the mean length of time employed by all women with one job by that of a random sample of 50 such women will be at most 20 weeks. My answer is:
(µ-20)- µ = 48-50 = -0.15373
13.01 13.01
(µ+2)-µ = 52-50 =0.15373
13.01 1301
Area less than z = 0.5596-0.4404 = 0.1192
The probability is approximately 0.1192 that the sampling error made in estimated the population mean of the length of time employed during the first 8 years of career will be at most 20 weeks.
Any help you can give me is very much appreciated.
Answer by jrfrunner(365)   (Show Source):
You can put this solution on YOUR website! If I understand your question, since its not real clear reading what you provide.
---
You have a skewed distribution for
X=time women with one job are employed during first 8 years of their career
--
X has a mean µ and standard deviation =92 (I dont see where you state the mean value)
--
The distribution of Xbar, "if taken from large samples, at least 30" will yield a normal distribution regardless of the original distribution, the larger the sample sizes the closer to a normal distribution with mean µ, standard error =std dev/sqrt(n)
--
In your case, if the sample is 50, then Xbar will have an approximate normal distribution with mean µ (again I dont see you saying what this is) and a standard error =92/sqrt(50)=13.01
--
You then want to know P(Xbar<20)
--
P(Xbar<20) =P(Z<(20-µ)/13.01) but I dont see where you show what µ is
if you know µ then substitute its value and compute Z, then use tables or excel or a statistcal calculator to get the answer the probablity.
|
|
|
