Q5: Suppose the amunt o fheating oil used annually by household in Iowa is norma

Question

Q5: Suppose the amunt o fheating oil used annually by household in Iowa is normally distributed with mean 200 gallons per household per year and standard deviation 40 gallons of heating oil per household per year.

(A) What is the probability a randomly selected sample of 50 Iowa household use more than 210 gallons of heating oil per year be?

(B) What is the probabilty that a randomly selected sample of 50 Iowa household use between 203 and 300 gallons per year be?

(C) If the memebers of a particular household were scared into using fuel conservation measure by newspaper accounts of the probable price of heating oil next year, and they decide they wanted to use less oil than 97.5% of all other Iowa households currently using heating oil. What is the maximum amount of oil they can use and still accomplish their conservation objective?

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    210      
u = mean =    200      
n = sample size =    50      
s = standard deviation =    40      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.767766953      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.767766953   ) =    0.038549936 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    203      
x2 = upper bound =    300      
u = mean =    200      
n = sample size =    50      
s = standard deviation =    40      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    0.530330086      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    17.67766953      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.702058455      
P(z < z2) =    1      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.297941545   [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.975 =   0.025      
          
Then, using table or technology,          
          
z =    -1.959963985      
          
As x = u + z * s,          
          
where          
          
u = mean =    200      
z = the critical z score =    -1.959963985      
s = standard deviation =    40      
          
Then          
          
x = critical value =    121.6014406 gallons   [ANSWER]  

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