Suppose a random variable is normally distributed. The probabilities for a85 and 142 are 10% and 65%, respectively. Find the mean and standard deviation to the nearest whole number.
Mean (Expected Value):
Mean(μ)=ΣxpMean (μ) = ΣxpMean(μ)=Σxp
Mean(μ)=(85×0.10)+(142×0.65)Mean (μ) = (85\times0.10)+(142\times0.65)Mean(μ)=(85×0.10)+(142×0.65)
Mean(μ)=8.50+92.30=100.80≈101Mean (μ) =8.50+92.30=100.80\approx101Mean(μ)=8.50+92.30=100.80≈101
Standard Deviation (σ):
σ=Σ(x−μ)2ρ\sigma=\sqrt{Σ(x-\mu)^{2} \rho}σ=Σ(x−μ)2ρ
σ=(85−101)2×0.1+(142−101)2×0.65\sigma=\sqrt{(85-101)^{2}\times0.1+(142-101)^2\times0.65}σ=(85−101)2×0.1+(142−101)2×0.65
σ=(256×0.1)+(1681×0.65)\sigma=\sqrt{(256\times0.1)+(1681\times0.65)}σ=(256×0.1)+(1681×0.65)
σ=25.6+168.1\sigma=\sqrt{25.6+168.1}σ=25.6+168.1
σ=193.7\sigma=\sqrt{193.7}σ=193.7
σ=13.9≈14\sigma=13.9\approx14σ=13.9≈14
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