Consumers spend an average of 1380 on a meal at a restaurant

Consumers spend an average of $13.80 on a meal at a restaurant in 2012. (Data extracted from www.alixpartners.com.) Assume that the amount spent on a restaurant meal is normally distributed and that the standard deviation is $2. What is the probability that a randomly selected person spent more than $15? What is the probability that a randomly selected person spent between $ 10 and $ 12? Between what two values will the middle 95% of the amounts spent fall?

Solution

A)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    15      
u = mean =    13.8      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    0.6      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.6   ) =    0.274253118 [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 =    10      
x2 = upper bound =    12      
u = mean =    13.8      
          
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.9      
z2 = upper z score = (x2 - u) / s =    -0.9      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.02871656      
P(z < z2) =    0.184060125      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.155343566   [ANSWER]

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C)

Note that              
              
Lower Bound = u - z(alpha/2) * s              
Upper Bound = u + z(alpha/2) * s              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025      
u = population mean =    13.8          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = population standard deviation =    2      
              
Thus,              
              
Lower bound =    9.880072031          
Upper bound =    17.71992797          
              
Thus, the confidence interval is              
              
(   9.880072031   ,   17.71992797   ) [ANSWER]