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Confidence Interval Estimate

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Added on  2019-09-26

Confidence Interval Estimate

   Added on 2019-09-26

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Individual AssignmentQuestion 1Part a)i) P(Z>−0.8)=0.788ii)P(−1.3<Z←0.7)=0.145iii)Z0.2=0.5793Part b)Z=x−μσ¿x−654i)P(x>70)=P(Z>70−654)¿P(Z>1.25)¿1−0.8944¿0.1056ii)P(x<60)=P(Z<60−654)¿P(Z←1.25)¿1−0.8944
Confidence Interval Estimate_1
¿0.1056iii)P(55<x<70)=P(55−654<Z<70−654)¿P(−2.5<Z<1.25)¿0.8944−0.0062¿0.8882Question 2Part a)i)Here,μ=100∧σ=20Therefore,z=96−10020=−0.2P( ́X<96)=P(z←0.2)¿0.4207ii)P(96< ́X<105)=P(96−10020<z<105−10020)¿P(−0.2<z<0.25)¿0.178
Confidence Interval Estimate_2
Part b)Z=x−μσ¿x−526i)P(x>60)=P(Z>60−526)¿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>60−52√12)¿P(Z>2.31)¿1−0.9896¿0.0104
Confidence Interval Estimate_3
Question 3Part a)Given ́X=500,σ=12,n=50i)α=0.05α2=0.025Zα2=1.96Confidence interval estimate of population mean ́X−Zα2∗σ√n, ́X+Zα2∗σ√n500−1.96∗12√50,500+1.96∗12√50496.674,503.326The 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 ́X−Zα2∗σ√n, ́X+Zα2∗σ√n
Confidence Interval Estimate_4

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