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BUS130 StatisticsAssignmentStudent Id & Name[Pick the date]

Part ANormal distributionAverage spending per weekμ= $21Standard deviationσ= $5(a)Probability that a randomly selected person would spend more than $25P(X>25)=?The requisite probability would be the area under the blue section of the curve.P(X>25)=P(X−μ>25−21)=P(X−μσ>25−215)Z=(X−μσ)=(25−215)=0.8P(X>25)=P(Z>0.8)= 1-P(Z<0.8)P(X>25)=1−0.7881=0.2119(¿ztable)1

Therefore, the probability thata randomly selected person would spend more than $25 is 0.2119.(b)Probability that a selected person would spend between $10 and $20P(10<X<20)=?The requisite probability would be the area under the blue section of the curve.P(10<X<20)=P(10−21<X−μ<20−21)¿P(10−215<X−μσ<20−215)Z=(X−μσ)=(10−215)=−2.2Z=(X−μσ)=(20−215)=−0.2P(10<X<20)=P(−2.2<Z←0.2)2

P(10<X<20)=P(−2.2<Z←0.2)=0.4068(¿ztable)Therefore, probability that a selected person would spend between $10 and $20 is 0.4068.(c)The middle 95% is indicated by the 0.95 area about the mean value. This represents that0.95/2 = 0.4750 area would fall on the left side of the mean and also the 0.4750 areawould fall on the right side of the mean.zvaluecoressponding¿95%confidenceinterval=1.96Hence,z=X−μσ1.96=(X−215)X=(−1.96∗5)+21X=−9.8+21PositivesignX=−9.8+21=11.8NegativesignX=9.8+21=30.8Therefore, 95% of the total amount of cash spent lie between the range $11.8 and $30.8.3

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