The weight of a small Starbucks coffee is a normally distributed random variable

Question

The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 315 grams and a standard deviation of 16 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.)

The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 315 grams and a standard deviation of 16 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.)

Explanation / Answer

Here mean=315 and sd=16

a. We need to find P(X>x)=0.20

So 0.5-P(0<X<x)=0.20

P(0<X<x)=0.30

Using z table we get z=0.8415 for which P(0<Z<0.8415)=0.30

z=x-mean/sd=x-315/16=0.8415

Hence x=16*0.8415+315= 328.464

b. We need to find P(x1<X<x2)=0.60

Using z table we get (-0.8415<z<0.8415)=0.60

z1=x-x1/sd=-0.8415

So x1=-0.8415*16+315= 301.536

X2=0.8415*16+315= 328.464

c. P(X>x)=0.80

P(0<X<x)=0.5-0.8=-0.3

Using z table we get P(0<z<-0.8415)=-0.3

So z=x-x/sd=-0.8415

So x=-0.8415*16+315= 301.536

d. P(X<x)=0.15

0.5+P(0<X<x)=0.15

P(0<X<x)=-0.35

Using z table we get z= -1.0365=x-mean/sd

So x=-1.0365*16+315= 298.416

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