Statistical Question.... Can anyone solve it?

At a large department store, the average number of years of service was 13.5 with a standard deviation of 6.2 years. If an employee is picked at random, what is the probability that the employee has worked for the company between 9 and 19 years?








possible solutions...








a. 0.3726





b. 0.4276





c. 0.5434





d. 0.5785





e. 0.6653





or none of these

Statistical Question.... Can anyone solve it?
For the large departmental store


Let x represent the number of years of service





Average number of years of service is 13.5 years


==%26gt; Mean (x bar) = 13.5 years





with a standard deviation of 6.2 years


==%26gt; SD (sigma x) = 6.2 years





Standard Normal Variate Z = (x - x bar)/(sigma x)


==%26gt; Z = (x - 13.5 years)/(6.2 years)





Probability that an employee selected has worked for the company between 9 and 19 years


==%26gt; P( 9 years %26lt; X %26lt; 19 years) = P(-0.73 %26lt; Z %26lt; 0.89)


[


At x = 9 years, Z = (9 years - 13.5 years)/(6.2 years)


==%26gt; Z = -4.5 years/6.2 years ==%26gt;Z= -0.73


At x = 19 years, Z = (19 years - 13.5 years)/(6.2 years)


==%26gt; Z = 5.5 years/6.2 years ==%26gt;Z= 0.89


]





==%26gt; P( 9 years %26lt; X %26lt; 19 years)


= P(-0.73 %26lt; Z %26lt; 0.89)


= P(-0.73 %26lt; Z %26lt; 0) + P( 0 %26lt; Z %26lt; 0.89)


= P(-0.73 %26lt; Z %26lt; 0) + P( 0 %26lt; Z %26lt; 0.89)


= P( 0 %26lt; Z %26lt; 0.73) + P( 0 %26lt; Z %26lt; 0.89)


[By the principle of symmetry]


=


(Area under the Standard Normal Curve (SNC) between Z = 0 and Z = %26lt; 0.73)


+ (Area under the SNC between Z = 0 and Z = %26lt; 0.89)





= 0.2673 + 0.3133


= 0.5806








http://www.futureaccountant.com/probabil...
Reply:i dunno