The amount of sodium in a randomly selected eight-ounce serving of chicken noodl

Question

The amount of sodium in a randomly selected eight-ounce serving of chicken noodle soup has a normal distribution with mean mu = 1737 mg and standard deviation sigma = 150 mg.^24 Suppose an eight-ounce serving is randomly selected. What is the probability that the amount of sodium is more than 1500 mg? What is the probability that the amount of sodium is between 1700 and 2000 mg? Find a symmetric interval about the mean such that 90% of all eight-ounce servings have amounts of sodium in that interval. Suppose a randomly selected eight-ounce serving has a sodium level of 2100 mg. Is there any evidence to suggest the mean sodium level reported above is false?

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1500      
u = mean =    1737      
          
s = standard deviation =    150      
          
Thus,          
          
z = (x - u) / s =    -1.58      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.58   ) =    0.942946567 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    1700      
x2 = upper bound =    2000      
u = mean =    1737      
          
s = standard deviation =    150      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.246666667      
z2 = upper z score = (x2 - u) / s =    1.753333333      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.402583103      
P(z < z2) =    0.960227596      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.557644494   [ANSWER]

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c)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.9      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.05      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.644853627      
By symmetry,          
z2 =    1.644853627      
          
As          
          
u = mean =    1737      
s = standard deviation =    150      
          
Then          
          
x1 = u + z1*s =    1490.271956      
x2 = u + z2*s =    1983.728044      

Hence, between 1490.27 and 1983.73. [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    2100      
u = mean =    1737      
          
s = standard deviation =    150      
          
Thus,          
          
z = (x - u) / s =    2.42      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.42   ) =    0.007760254

Yes, because this is a very small probability. [ANSWER]

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