A sample of 9 customers in dollars spent for lunch for fastf

A sample of 9 customers in dollars spent for lunch for fast-food:

4.20 5.03 5.89 6.45 7.38 7.54 8.46 8.47 9.87

a. Compute the mean and median.

b. Compute the variance, standard deviation, range, and co-efficient of variation.

c. Are the data skewed? How?

d. Based on the results of a through c, what conclusions can you reach concerning the amount customers spent on lunch?

Solution

a)

Getting the mean, X,          
          
X = Sum(x) / n          
Sum(x) =    63.29      
          
Thus,          
X =    7.032222222   [answer, mean]

Ordering them,

4.2
5.03
5.89
6.45
7.38
7.54
8.46
8.47
9.87

Thus, the median is the 5th term,

median = 7.38 [answer, median]

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Setting up tables,          
x   x - X   (x - X)^2  
4.2   -2.832222222   8.021482716  
5.03   -2.002222222   4.008893827  
5.89   -1.142222222   1.304671605  
6.45   -0.582222222   0.338982716  
7.38   0.347777778   0.120949383  
7.54   0.507777778   0.257838272  
8.46   1.427777778   2.038549383  
8.47   1.437777778   2.067204938  
9.87   2.837777778   8.052982716  
          
          
Thus, Sum(x - X)^2 =    26.21155556      
          
Thus, as           
          
s^2 = Sum(x - X)^2 / (n - 1)          
          
As n =    9      
          
s^2 =    3.276444444   [answer, variance]
          
Thus,          
          
s =    1.810095148   [answer, standard deviation]

Also, we have      
      
Maximum =    9.87  
Minimum =    4.2  
Range = max - min =    5.67   [answer, range]

coefficient of variation = s / mean = 0.257400163 [answer]

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c) slightly, as median > mean. Hence, it is skewed to the left.

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d)

The mean amount spent is $7.03, but the median amount is $7.38, which indicates that more people spend amount greater than $7.03 than below 7.03.