1 The weight of turkeys is normally distributed with a mean

1. The weight of turkeys is normally distributed with a mean of 22 pounds and a standard deviation of 5 pounds.

a. Find the probability that a randomly selected turkey weighs between 20 and 26 pounds. Your answer is . _______Round to 3 decimals and keep \'0\' before the decimal point.

b. Find the probability that a randomly selected turkey weighs below 12 pounds. Your answer is _________. Round to 3 decimals and keep \'0\' before the decimal point.

Solution

a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    20      
x2 = upper bound =    26      
u = mean =    22      
          
s = standard deviation =    5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.4      
z2 = upper z score = (x2 - u) / s =    0.8      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.344578258      
P(z < z2) =    0.788144601      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.443566343 = 0.444 [ANSWER]      

b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    12      
u = mean =    22      
          
s = standard deviation =    5      
          
Thus,          
          
z = (x - u) / s =    -2      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -2   ) =    0.022750132 = 0.023 [ANSWER]