# SOLUTION: Having a lot of trouble with this. Please help and show me how so I can actually learn from all this. Thanks! PS- I am allowed to use Megastat on this so any direction there wou

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: Having a lot of trouble with this. Please help and show me how so I can actually learn from all this. Thanks! PS- I am allowed to use Megastat on this so any direction there wou      Log On

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 Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Probability-and-statistics Question 139732: Having a lot of trouble with this. Please help and show me how so I can actually learn from all this. Thanks! PS- I am allowed to use Megastat on this so any direction there would be welcomed as well. The bi-monthly starting salaries of recent statistician graduates follow the normal distribution with a mean of \$2,625 and a standard deviation of \$350. (shos all your work please) A. What is the z-value for a salary of \$2,200? B. What is the approximate percent of statisticians making between \$2,625 & \$2,975? C. What is the approximate percent of statisticians making between \$2,275 & \$2,625?Answer by stanbon(57387)   (Show Source): You can put this solution on YOUR website!The bi-monthly starting salaries of recent statistician graduates follow the normal distribution with a mean of \$2,625 and a standard deviation of \$350. (shos all your work please) A. What is the z-value for a salary of \$2,200? z(2200) = (2200-2625)/350 = -1.21428... ----------------------------------------------------- B. What is the approximate percent of statisticians making between \$2,625 & \$2,975? Since the mean is 2625 the z-value of 2625 is zero z(2975) = (2975-2625)/350 = 1 P(2625 < x < 2975) = P(0 < z < 1) = 0.3413.. -------------------------------- C. What is the approximate percent of statisticians making between \$2,275 & \$2,625? P( 2275 < x < 2625) = P(-1 < z < 0) = 0.3413 ================== Cheers, Stan H.