document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=346588>Question 346588</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can you please tell me if my answers are correct:<BR>\n" );
document.write( "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:<BR>\n" );
document.write( "a. Determine the sampling distribution of the sample mean for simple random samples of 50 women with one job. my answer is 92/&#8730;50=13.01 – sampling distribution of the sample mean will be approximately normal because the sample is large enough.<BR>\n" );
document.write( "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: <BR>\n" );
document.write( "(µ-20)- µ  = 48-50 = -0.15373<BR>\n" );
document.write( "13.01           13.01\r<BR>\n" );
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document.write( "(µ+2)-µ  = 52-50  =0.15373<BR>\n" );
document.write( "13.01         1301<BR>\n" );
document.write( "Area less than z = 0.5596-0.4404 = 0.1192<BR>\n" );
document.write( "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.\r<BR>\n" );
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document.write( "Any help you can give me is very much appreciated.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #247793 by </B><A HREF=/tutors/aboutme.mpl?userid=jrfrunner><B>jrfrunner(365)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jrfrunner><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=jrfrunner><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=247793\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If I understand your question, since its not real clear reading what you provide.<BR>\n" );
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document.write( "You have a skewed distribution for<BR>\n" );
document.write( " X=time women with one job are employed during first 8 years of their career <BR>\n" );
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document.write( "X has a mean µ and standard deviation =92  (I dont see where you state the mean value)<BR>\n" );
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document.write( "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)<BR>\n" );
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document.write( "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<BR>\n" );
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document.write( "You then want to know P(Xbar<20)  <BR>\n" );
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document.write( "P(Xbar<20) =P(Z<(20-µ)/13.01)  but I dont see where you show what µ is <BR>\n" );
document.write( "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. </SPAN>\n" );
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