The weight of a small Starbucks coffee is a normally distrib

The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 340 grams and a standard deviation of 11 grams. Find the weight that corresponds to each event. (Use Excel or Appendix C for calculation of z-value. Round your final answers to 2 decimal places.)

Solution

(a) Highest 30 percent

P(X>x)=0.3

--> P((X-mean)/s <(x-340)/11) =1-0.3=0.7

--> P(Z<(x-340)/11) =0.7

--> (x-340)/11 =0.52 (from standard normal table)

So x= 340+0.52*11=345.72

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(b)Middle 70 percent

P(X<x)=(1-0.7)/2 =0.15

--> P(Z<(x-340)/11) =0.15

--> (x-340)/11= -1.04 (from standard normal table)
So x= 340-1.04*11=328.56

P(X<x)=0.85

--> P(Z<(x-340)/11) =0.85

--> (x-340)/11= 1.04 (from standard normal table)
So x= 340+1.04*11=351.44

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(c)Highest 90 percent

P(X<x)=0.9

--> P(Z<(x-340)/11) =0.9

--> (x-340)/11= 1.28 (from standard normal table)

So x= 340+1.28*11=354.08

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(d)Lowest 20 percent

P(X<x)=0.2

--> P(Z<(x-340)/11) =0.2

--> (x-340)/11 = -0.84 (from standard normal table)

So x= 340-0.84*11=330.76