Question 1088546
A manufacturer makes chocolate bars with a mean weight of 110 grams and a standard
deviation of 2 grams. The weight is normally distributed.
(a) What proportion of the bars are likely to be less than 106 grams in weight?(
z(106) = (106-110)/2 = -4/2 = -2
P(x < 106) = P(z < -2) = normalcdf(-100,-2) = 0.0228 or 2.28%
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(b) The manufacturer decides to make bigger bars with a mean weight of 115 grams with the same standard deviation as before.
What mean weight will have to be aimed at if no more than one bar in 100 is less than 115 grams in weight? (The weight of the bigger bars is also normally
distributed).
Find the z-value with a left-tail of 1%
invNorm(0.01) = -2.3263
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Solve for mean::
-2.3263 = (115-u)/3
-6.9790 = 115-u
u = 115+6.9790
u = 121.979
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Cheers,
Stan H.
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