This exercise uses the normal probability density function and requires the use

Question

This exercise uses the normal probability density function and requires the use of either technology or a table of values of the standard normal distribution. The cash operating expenses of the regional phone companies during the first half of 1994 were distributed about a mean of $29.98 per access line per month, with a standard deviation of $2.15. Company A's operating expenses were $29.00 per access line per month. Assuming a normal distribution of operating expenses, estimate the percentage of regional phone companies whose operating expenses were closer to the mean than the operating expenses of Company A were to the mean. (Round your answer to two decimal places.) %

Explanation / Answer

Here, company A's z score is, as z = (x-u)/sigma,

zA = (29.00 - 29.98)/2.15 = -0.455813953

Hence, those close to the mean than company A have z scores from -0.455813953 to 0.455813953.

z1 = lower z score =    -0.455813953      
z2 = upper z score =     0.455813953      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.324261885      
P(z < z2) =    0.675738115      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.35147623 = 35.147623% [ANSWER]      

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