14. 0/6.66 points | Previous Answers WaneFMAC6 P.2.039. My No This exercise uses

Question

14. 0/6.66 points | Previous Answers WaneFMAC6 P.2.039. My No 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 $29.92 per access line per month, with a standard deviation of $2.15. Company A's operating expenses were $28.00 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.) 31.33 x%

Explanation / Answer

here as diference betwene operating expense and mean expense =29.92-28 =1.92

hence probability =P(-1.92/2.15<Z<1.92/2.15)=P(-0.8930<Z<0.8930)=0.8141-0.1859=06282 ~ 62.82%

( if we round z values to 2 decimal places ; probability =62.65%)

I think you have calculated it right only has to multiply by 2 so that get both side probability

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