Chocolate bars produced by a certain machine are labelled with 100 g. The distri

Question

Chocolate bars produced by a certain machine are labelled with 100 g. The distribution of actual weights of these chocolate bars is Normal with a mean of 103 g and standard deviation of 1.8 g. A chocolate bar is considered underweight if it weighs less than 100 g. Define the appropriate random variable State the distribution of the random variable (In proper distribution notation) What proportion of chocolate bar weigh between than 100 g? What proportion of chocolate bars weigh between 104 and 106 g? How should the chocolate bar wrappers be labelled (number of grams on the label) so that only 1% of such bars are underweight?

Explanation / Answer

2.

a)

Weights of chocolate bars

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b)

It is normally distrbuted with mean 103 g and standard deviation of 1.8 g.

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    100      
u = mean =    103      
          
s = standard deviation =    1.8      
          
Thus,          
          
z = (x - u) / s =    -1.666666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.666666667   ) =    0.047790352 [ANSWER]

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d)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    104      
x2 = upper bound =    106      
u = mean =    103      
          
s = standard deviation =    1.8      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.555555556      
z2 = upper z score = (x2 - u) / s =    1.666666667      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.710742639      
P(z < z2) =    0.952209648      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.241467008   [ANSWER]

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e)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.01      
          
Then, using table or technology,          
          
z =    -2.326347874      
          
As x = u + z * s,          
          
where          
          
u = mean =    103      
z = the critical z score =    -2.326347874      
s = standard deviation =    1.8      
          
Then          
          
x = critical value =    98.81257383   [ANSWER]  
     

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