Consider the population of all 1gal cans of dusty rose paint

Consider the population of all 1-gal cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean = 5 ml and standard deviation = 0.2 ml is a reasonable model for the distribution of the variable x = amount of red dye in the paint mixture. Use the normal distribution model to calculate the probabilities below. (Round all answers to four decimal places.)

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    5      
u = mean =    5      
          
s = standard deviation =    0.2      
          
Thus,          
          
z = (x - u) / s =    0      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0   ) =    0.5 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    5.4      
u = mean =    5      
          
s = standard deviation =    0.2      
          
Thus,          
          
z = (x - u) / s =    2      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   2   ) =    0.977249868 [ANSWER]

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c)

It is just the same as above, as points do not contribute to the total area under the curve,

P(x<=5.4) = 0.977249868 [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 =    4.6      
x2 = upper bound =    5.2      
u = mean =    5      
          
s = standard deviation =    0.2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.818594614   [ANSWER]

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e)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    4.5      
u = mean =    5      
          
s = standard deviation =    0.2      
          
Thus,          
          
z = (x - u) / s =    -2.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -2.5   ) =    0.993790335 [ANSWER]

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F)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    4      
u = mean =    5      
          
s = standard deviation =    0.2      
          
Thus,          
          
z = (x - u) / s =    -5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -5   ) =    0.999999713 [ANSWER]