Given that,μ=62,σ=3.5,Let Z1=X−623.5,thenP(Z>Z1)=0.7580.5+P(−Z1<Z<0)=0.758P(−Z1<Z<0)=0.258−Z1=0.7 ⟹ Z1=−0.7∴−0.7=X−623.5X=(−0.7×3.5)+62=59.55the minimum specification is 59.55 inchesGiven \; that, μ=62, σ=3.5,\\ Let\; Z_{1}=\frac{X-62}{3.5}, then\\ P(Z>Z_{1})=0.758\\ 0.5+P(-Z_{1}<Z<0)=0.758\\ P(-Z_{1}<Z<0)=0.258\\ -Z_{1}=0.7 \implies Z_{1}=-0.7\\ \therefore -0.7=\frac{X-62}{3.5}\\ X=(-0.7\times 3.5)+62=59.55\\ \text{the minimum specification is}\;59.55\; inchesGiventhat,μ=62,σ=3.5,LetZ1=3.5X−62,thenP(Z>Z1)=0.7580.5+P(−Z1<Z<0)=0.758P(−Z1<Z<0)=0.258−Z1=0.7⟹Z1=−0.7∴−0.7=3.5X−62X=(−0.7×3.5)+62=59.55the minimum specification is59.55inches
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