M&M plain candies have a mean weight of 0.8565 g and a standard deviation of 0.0

Question

M&M plain candies have a mean weight of 0.8565 g and a standard deviation of 0.0518g. The M&M candies used in Data Set 20 came from a package containing 465 candies, and the package label stated that the net weight is 396.9g
1. If on M&M plain candy is randomly seleced, find the probability that it weighs between 0.8535g and 0.8565g
2. If one M&M plain candy is randomly selected, find the probability that it weighs more than 0.8535g
3. If package (Containing 465 M&M candies) is randomly selected, find the probability its mean weight is between 0.8535g and 0.8565g
4. If 465 M&M plain candies are randomly selected, find the probability their mean weight is at least 0.8535g?
5. Does it seem that the Mars cpompany is providing M&M consumers with the amount claimed on the label? (If every package has 465 candies, the mean weight of the candies must exceed 396.9/465 = 08535g for the net contents to weigh atleast 396.9g)

(Answer with explanation please)

Explanation / Answer

1.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.8535      
x2 = upper bound =    0.8565      
u = mean =    0.8565      
          
s = standard deviation =    0.0518      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.057915058      
z2 = upper z score = (x2 - u) / s =    0      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.476908144      
P(z < z2) =    0.5      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.023091856   [ANSWER]

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2.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.8535      
u = mean =    0.8565      
          
s = standard deviation =    0.0518      
          
Thus,          
          
z = (x - u) / s =    -0.057915058      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -0.057915058   ) =    0.523091856 [ANSWER]

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3.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    0.8535      
x2 = upper bound =    0.8565      
u = mean =    0.8565      
n = sample size =    465      
s = standard deviation =    0.0518      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.248872123      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.105855925      
P(z < z2) =    0.5      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.394144075   [ANSWER]

*******************

4.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    0.8535      
u = mean =    0.8565      
n = sample size =    465      
s = standard deviation =    0.0518      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.248872123      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.248872123   ) =    0.894144075 [ANSWER]

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5.

It seems so, because part 4 is really a high probability. [ANSWER]
  

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