[college intro statistics] simple problems. please answer #5 and #6. Show your w

Question

[college intro statistics] simple problems. please answer #5 and #6. Show your work in detail thanks!

5. (10 pts) Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Suppose that the actual weights of theses candy bars vary according to a normal distribution with mean =2.2 ounces and standard deviation =0.06 ounces. If a packages that contains 5 of these candy bars is chosen randomly, what is the probability that at least one of the candy bars weights less than the advertised weight?

Explanation / Answer

6. X1~N(40,10) and X2~N(50,12) independently.

hence X1+X2 being a linear combination of normal variates would also follow a normal distribution with

mean=E[X1+X2]=E[X1]+E[X2]=40+50=90

and variance=V[X1+X2]=V[X1]+V[X2] [since they are independent , there is no covariance term]

=10+12=22

so P[X1+X2<=80]=P[[(X1+X2)-90]/sqrt(22)<=(80-90)/sqrt(22)]=P[Z<=-2.132] Z~N(0,1)

=0.01650342 [answer]

5. the wrapper of a certain candy bar lists its weight as 2.13 ounces

let X be the random variable denoting the weights of the candy bars.

by question X~N(2.2,0.062)

let p be the probability that a randomly selected candy bar weighs less than 2.13 ounce

so p=P[X<=2.13]=P[(X-2.2)/0.06<(2.13-2.2)/0.06]=P[Z<=-1.1667]=0.1216658

let Y denotes the number of candy bars weighs less than 2.13 ounces out of 5 bars

then Y~Bin(5,0.1216658)

hence the probability that at least one of the 5 candy bars weigh less than 2.13 ounces is

P[Y>=1]=1-P[Y=0]=1-5C00.12166580(1-0.1216658)5-0=0.477244 [answer]

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