A normal distribution has the mean of 150.5. if 70.54% of the area under the curve lies to the left of 177.5 find a. the standard deviation?
P(x<177.5)=0.7054P(x<177.5)=0.7054P(x<177.5)=0.7054
z=0.54z=0.54z=0.54
z=x−μσz=\frac{x-\mu}{\sigma}z=σx−μ
the standard deviation:
σ=x−μz=177.5−150.50.54=50\sigma=\frac{x-\mu}{z}=\frac{177.5-150.5}{0.54}=50σ=zx−μ=0.54177.5−150.5=50
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