for normal distribution z score =(X-μ)/σx | |
here mean= μ= | 180 |
std deviation =σ= | 17.000 |
a)
probability =P(X>197)=P(Z>(197-180)/17)=P(Z>1)=1-P(Z<1)=1-0.8413=0.1587 |
(
if using excel use command :1-norm.dist(197,180,17,true) |
b)
probability =P(180<X<197)=P((180-180)/17)<Z<(197-180)/17)=P(0<Z<1)=0.8413-0.5=0.3413 |
if using excel use command :norm.dist(197,180,17,true)-norm.dist(180,180,17,true) |
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