Question

Assume that the amount of cornflakes in a box is normally distributed with a mean of 16 oz. and a standard deviation of 0.1 oz.

a) Determine the percent of boxes that will contain between 15.83 oz. and 16.32 oz. of cornflakes?

b) Determine the percent of boxes that will contain more than 16.16 oz. of cornflakes.

c) If the manufacturer produces 300,000 boxes, how many of them will contain less than 15.83 oz. of cornflakes?

d) If the manufacturer produces 300,000 boxes, how many of them will contain more than 16.16 oz. of cornflakes?

Answer 1

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The normal distribution parameters are given as:

Mean = 16 oz
Stdev = .1 oz

We will these parameters along with the standardization formula to solve the problem. The formula for standardization is : Z = (X-Mean)/Stdev

a.P(15.83<Z<16.32)
standardizing using above formula:
= P( (15.83-16)/.1 < Z < (16.32-16)/.1)
= P(-1.7 < Z < 3.2)
= P(Z<3.2 ) - P(Z<-1.7) [use the Z-tables to convert the Z to cumulative probabilities]
= 0.9993 - 0.0446  
= 0.9547

b. P(X>16.16)
standardizing using above formula:
P(Z> (16.16-16)/.1)
= P(Z>1.6) [use the Z-tables to convert the Z to cumulative probabilities]
= 1-P(Z<=1.6)
= 1- 0.9452
= 0.0548

c. This will be
= number of total boxes manufactured*P(X<15.83) = 300000*P(X<15.83)
standardizing using above formula:
P(Z> (15.83-16)/.1)
= P(Z>-1.7)
= 1 - P(Z<=-1.7) [use the Z-tables to convert the Z to cumulative probabilities]
= 1- 0.044565
= 0.955435

So, Answer is 300000*.955435 = 286630.36 or 286630

Answer: 286630

d.

This will be 300,000*P(X>16.16)
300000*P(Z> (16.16 -16)/.1)
= 300000*P(Z>1.6)
= 300000*(1-P(Z<=1.6) [use the Z-tables to convert the Z to cumulative probabilities]
= 300000*(1-0.945201)
= 16439.7 or 16440

Answer: 16440