SOLUTION: CLAIMS FILED DURING YEAR FOR INSURANCE HAD MEAN OF NORMAL DISTRIBUTION =$4560.STD DEVIATION OF NORMAL DISTRIB WAS $1500. A SAMPLE OF 100 CLAIMS WAS SELECTED. FIND PROBABILITY THAT

Question 691129: CLAIMS FILED DURING YEAR FOR INSURANCE HAD MEAN OF NORMAL DISTRIBUTION =$4560.STD DEVIATION OF NORMAL DISTRIB WAS $1500. A SAMPLE OF 100 CLAIMS WAS SELECTED. FIND PROBABILITY THAT THE MEAN CLAIM FOR THE 100 SAMPLES WAS BETWEEN 4200 AND 4600 DOLLARS AND PROB IT WAS LESS THEN 4000 DOLLARS. PLEASE HELP ME
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
CLAIMS FILED DURING YEAR FOR INSURANCE HAD MEAN OF NORMAL DISTRIBUTION =$4560.STD DEVIATION OF NORMAL DISTRIB WAS $1500. A SAMPLE OF 100 CLAIMS WAS SELECTED. FIND PROBABILITY THAT THE MEAN CLAIM FOR THE 100 SAMPLES WAS BETWEEN 4200 AND 4600 DOLLARS AND PROB IT WAS LESS THEN 4000 DOLLARS. PLEASE HELP ME
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std of means of samples of size n = 100 is 1500/sqrt(100) = 150
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z(4600) = (4600-4560)/150 = 40/150 = 4/15
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z(4200) = (4200-4560)/150 = -260/150 = -26/15
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P(4200<= x-bar <= 4600) = P(-26/15<= z <=4/15)
= normalcdf(-26/15,4/15)
= 0.5636
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Cheers,
Stan H.
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