At your supermarket the typical shopper spends 186 with a st

At your supermarket, the typical shopper spends $18.6 with a standard deviation of $12.00. You are wondering what would happen in a typical morning hour with 400 typical shoppers, assuming that each one shops independently. a) What is the probability that the mean amount spent by the 400 shoppers is at least $18.00? b) Did you need to make any assumptions in order to answer part a)? Why or why not? Explain. c) Find the 30th percentile for the average amount spent by the 400 shoppers

Solution

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    18      
u = mean =    18.6      
n = sample size =    400      
s = standard deviation =    12      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1   ) =    0.841344746 [ANSWER]

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b)

No, because by central limit theorem, any distirbution will have an approximately normal distribution of sample means.

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.3      
          
Then, using table or technology,          
          
z =    -0.524400513      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    18.6      
z = the critical z score =    -0.524400513      
s = standard deviation =    12      
n = sample size =    400      
Then          
          
x = critical value =    18.28535969   [ANSWER]