Business Statistics Class At the end of a festival the organ

Business Statistics Class

At the end of a festival the organizers estimated that a family of participants spent in average $40.00 with a standard deviation of $8.00. If 49 participants (49 = size of the sample) are selected randomly, what’s the likelihood that their mean spent amount will be within $3 of the population mean? (mean +/- 3; between $37 and $43)

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    37      
x2 = upper bound =    43      
u = mean =    40      
n = sample size =    49      
s = standard deviation =    8      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.625      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.625      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.004332448      
P(z < z2) =    0.995667552      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.991335103   [ANSWER]