### BUS130 Statistics Assignment

Added on - 10 May 2020

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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(Xμ>2521)=P(Xμσ>25215)Z=(Xμσ)=(25215)=0.8P(X>25)=P(Z>0.8)= 1-P(Z<0.8)P(X>25)=10.7881=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(1021<Xμ<2021)¿P(10215<Xμσ<20215)Z=(Xμσ)=(10215)=2.2Z=(Xμσ)=(20215)=0.2P(10<X<20)=P(2.2<Z0.2)2
P(10<X<20)=P(2.2<Z0.2)=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=Xμσ1.96=(X215)X=(1.965)+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  