a population size N = 150 has μ = 8 and standard deviation of 5.4. What is the probability that a random size n=20 will have a mean of 9.5 above
Given,
Mean of population μ=8\mu=8μ=8
Standard deviation σ=5.4\sigma = 5.4σ=5.4
Using
P(Xˉ>9.5)=Xˉ−μσnP(\bar X >9.5)= \dfrac{\bar X-\mu}{\frac{\sigma}{\sqrt n}}P(Xˉ>9.5)=nσXˉ−μ
P(Xˉ>9.5)=9.5−85.420=1.51.20=1.25P( \bar X>9.5)= \dfrac{9.5-8}{\frac{5.4}{\sqrt{20}}}=\dfrac{1.5}{1.20}=1.25P(Xˉ>9.5)=205.49.5−8=1.201.5=1.25
So, P(Z>1.25)=0.1056P(Z>1.25)=0.1056P(Z>1.25)=0.1056
Hence, Probability P(Z>1.25)= 0.1056
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