The mean amount purchased by a typical customer at Churchill

The mean amount purchased by a typical customer at Churchill\'s Grocery Store is $21.00 with a standard deviation of $7.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 42 customers, answer the following questions. a. What is the likelihood the sample mean is at least $22.50? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability b. What is the likelihood the sample mean is greater than $20.00 but less than $22.50? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability c. Within what limits will 99 percent of the sample means occur? (Round your answers to 2 decimal places.) Sample mean and

Solution

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    22.5      
u = mean =    21      
n = sample size =    42      
s = standard deviation =    7      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.39      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.39   ) =    0.0823 [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 =    20      
x2 = upper bound =    22.5      
u = mean =    21      
n = sample size =    42      
s = standard deviation =    7      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.93      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.39      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.1762      
P(z < z2) =    0.9177      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.7415   [ANSWER]

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c)

These correspond to the endpoints with left tailed area of 0.005 and 0.995.

First, we get the z score from the given left tailed area. As          

Left tailed area =    0.005      
          
Then, using table or technology,          
          
z =    -2.58      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    21      
z = the critical z score =    -2.58      
s = standard deviation =    7      
n = sample size =    42      
Then          
          
x = critical value =    18.2132815      
  

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First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.995      
          
Then, using table or technology,          
          
z =    2.58      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    21      
z = the critical z score =    2.58      
s = standard deviation =    7      
n = sample size =    42      
Then          
          
x = critical value =    23.7867185      

Thus, it is between 18.21 and 23.79. [ANSWER]