Hi

mean of 1,600 hours and a standard deviation of 400 hours.

sample 64;  s = 400/sqrt(64) = 50

19)P(x > 1550),   z = (1550 - 1600)/50 = -50/50 = -1

P(x > 1550) = P(z > -1) = 1 -.1587(area to the left) =  .8413  0r 84.13%



20) Referring to Table 7-4, the probability is 0.15 that the sample mean lifetime is more than how many hours?

z =invNorm(.85) = 1.0364 

1.0364 = (X - 1600)/50 ,  50*1.0364 + 1600 = 51.82+ 1600 = 1652



Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  

z-values indicative of the Area to the left under the curve.

Note: z = 0 (x value the mean) 50% of the area under the curve is to the left and %50 to the right



 

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