The average amount parents spent per child on backtoschool i

The average amount parents spent per child on back-to-school items in August 2012 was $650 with a standard deviation of $245. Assume the amount spent on back-to-school items is normally distributed. The probability is 0.27 that the amount spent on a randomly selected child will be between what two values equidistant from the mean? Find the lower endpoint and the upper end point.

Solution

If the middle area is 0.27, the the sum of the outer areas is 1-0.27 = 0.73.

Thus, the left tailed area of the lower end is 0.73/2 = 0.365.

Thus, for lower end:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.365      
          
Then, using table or technology,          
          
z =    -0.345125531      
          
As x = u + z * s,          
          
where          
          
u = mean =    650      
z = the critical z score =    -0.345125531      
s = standard deviation =    245      
          
Then          
          
x = critical value =    565.4442448   [LOWER END]

For higher end, the left tailed area is 0.365 + 0.27 = 0.635.

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.635      
          
Then, using table or technology,          
          
z =    0.345125531      
          
As x = u + z * s,          
          
where          
          
u = mean =    650      
z = the critical z score =    0.345125531      
s = standard deviation =    245      
          
Then          
          
x = critical value =    734.5557552   [UPPER END]

Thus, the lower and upper endpoints are 565.4442448 and 734.5557552 [ANSWERS]