Question 172512
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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Find the z-value of 73.5:
z(73.5) = (73.5-72.5)/4 = 0.25
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Then P(x>73.5) = P(z>0.25) = 0.4013
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Cheers,
Stan H.