Question on normal distribution
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Question on normal distribution

[From: ] [author: ] [Date: 11-07-01] [Hit: ]
-What proportion of drivers in this category is supposed to have their monthly premium above $80?-What proportion of drivers in this category is supposed to have their monthly premium below $70?-What proportion of drivers in this category is supposed to have their monthly premium within the interval from $72 to $78?-What proportion of drivers in this category is supposed to have their premium within the interval from $74 to $84?P (x > 80) = P (z > 0.75) = 0.......
Monthly car insurance premium for a certain category of drivers is assumed to have the normal distribution, with the population mean of $74 and population standard deviation of $8.

-What proportion of drivers in this category is supposed to have their monthly premium above $80?
-What proportion of drivers in this category is supposed to have their monthly premium below $70?
-What proportion of drivers in this category is supposed to have their monthly premium within the interval from $72 to $78?
-What proportion of drivers in this category is supposed to have their premium within the interval from $74 to $84?

-
z = ( x - μ ) / σ
P (x > 80) = P (z > 0.75) = 0.2266
P ( x < 70) = P ( z < -0.5 ) = 0.3085
P (72 < x < 78) = P (-0.25 < z < 0.5) = 0.2902
P (74 < x < 84) = P (0 < z < 1.25) = 0.3944
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