# SOLUTION: A TEST IS NORMALLY DISTRIBUTED WITH A MEAN OF 72.5 AND A STANDARD DEVIATION OF 4, WHAT IS THE PROBABILITY OF GETTING MORE THAN73.5 IN THIS TEST?

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 Click here to see ALL problems on Probability-and-statistics Question 172512This question is from textbook ELEMENTARY STATISTICS : A TEST IS NORMALLY DISTRIBUTED WITH A MEAN OF 72.5 AND A STANDARD DEVIATION OF 4, WHAT IS THE PROBABILITY OF GETTING MORE THAN73.5 IN THIS TEST?This question is from textbook ELEMENTARY STATISTICS Answer by stanbon(60777)   (Show Source): You can put this solution on YOUR website!A TEST IS NORMALLY DISTRIBUTED WITH A MEAN OF 72.5 AND A STANDARD DEVIATION OF 4, WHAT IS THE PROBABILITY OF GETTING MORE THAN73.5 IN THIS TEST? ------- Find the z-value of 73.5: z(73.5) = (73.5-72.5)/4 = 0.25 -------------------------------- Then P(x>73.5) = P(z>0.25) = 0.4013 ----------------------- Cheers, Stan H.