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# Confidence Interval Estimate

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Individual AssignmentQuestion 1Part a)i) P(Z>0.8)=0.788ii)P(1.3<Z0.7)=0.145iii)Z0.2=0.5793Part b)Z=xμσ¿x654i)P(x>70)=P(Z>70654)¿P(Z>1.25)¿10.8944¿0.1056ii)P(x<60)=P(Z<60654)¿P(Z1.25)¿10.8944 ¿0.1056iii)P(55<x<70)=P(55654<Z<70654)¿P(2.5<Z<1.25)¿0.89440.0062¿0.8882Question 2Part a)i)Here,μ=100σ=20Therefore,z=9610020=0.2P( ́X<96)=P(z0.2)¿0.4207ii)P(96< ́X<105)=P(9610020<z<10510020)¿P(0.2<z<0.25)¿0.178 Part b)Z=xμσ¿x526i)P(x>60)=P(Z>60526)¿P(Z>1.33)¿0.0918ii)Let X̅ denote the mean amount of work per week for three randomly selected professors.Since,XN(52,62)P( ́X>60)=P(Z>605212)¿P(Z>2.31)¿10.9896¿0.0104 Question 3Part a)Given ́X=500,σ=12,n=50i)α=0.05α2=0.025Zα2=1.96Confidence interval estimate of population mean ́XZα2σn, ́X+Zα2σn5001.961250,500+1.961250496.674,503.326The lower and upper confidence limits are LCL = 496.674 and UCL =503.326 respectively.ii)α=0.01α2=0.005Zα2=2.575Confidence interval estimate of population mean ́XZα2σn, ́X+Zα2σn ## Found this document preview useful?

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