A box of tickets has an average of 100 and an SD of 20 Four

A box of tickets has an average of 100 and an SD of 20. Four hundred draws will be made at random with replacement from this box.

a. Estimate the chance that the average of the draws will be in the range 80 to 120

b. Estimate the chance that the average of the draws will be in the range 99 to 101

Solution

a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    80      
x2 = upper bound =    120      
u = mean =    100      
n = sample size =    400      
s = standard deviation =    20      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -20      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    20      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0      
P(z < z2) =    1      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    1   [or, very very close to 1]

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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 =    99      
x2 = upper bound =    101      
u = mean =    100      
n = sample size =    400      
s = standard deviation =    20      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.682689492   [ANSWER]