I have a bellshaped distribution with a mean of 31 pounds an

I have a bell-shaped distribution with a mean of 31 pounds and a standard deviation of 4 pounds. a,) approximately what percentage of the population will be between 27 pounds and 43 pounds?, b) between 23 pounds and 35 pounds?, c) below 35 pounds?, d) between 19 and 39 pounds, e) under 23 pounds?, f) more than 43 pounds?, g) between 23 and 27 pounds?

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    27      
x2 = upper bound =    43      
u = mean =    31      
          
s = standard deviation =    4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    3      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.998650102      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.839994848 = 83.9994848%   [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 =    23      
x2 = upper bound =    35      
u = mean =    31      
          
s = standard deviation =    4      
          
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 = 81.8594614% [ANSWER]

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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 =    35      
u = mean =    31      
          
s = standard deviation =    4      
          
Thus,          
          
z = (x - u) / s =    1      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   1   ) =    0.841344746 = 84.1344746% [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 =    19      
x2 = upper bound =    39      
u = mean =    31      
          
s = standard deviation =    4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -3      
z2 = upper z score = (x2 - u) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.001349898      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.97589997 = 97.589997% [ANSWER]

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