A certain trail-mix recipe calls for three protein-containing ingredients. The f

Question

A certain trail-mix recipe calls for three protein-containing ingredients. The first has a protein content that is normally distributed with a mean of BO and a standard deviation of 6. The second has a protein content that is normally distributed with a mean of 10 and a standard deviation of 4, and the third has a protein content that is normally distributed with a mean of 5 and a standard deviation of 2. If the recipe calls for 25% of ingredient 1, 10% of ingredient 2 and 20% of ingredient 3, what is the probability that the total protein content in the trail-mix is more than 10 (percent), i.e. find P(0.25X_1 + 0.1X_2 + 0.2X_3 > 12)

Explanation / Answer

here mean of 0.25X1+0.1X2+0.2X3=0.25*30+0.1*10+0.2*5=9.5

and std deviation =((0.25*6)2+(4*0.1)2+(2*0.2)2)1/2 =1.603

hence P(0.25X1+0.1X2+0.2X3>12)=1-P(Z<(12-9.5)/1.603)=1-P(Z<1.5595)=1-0.9406=0.0594

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