Acivil service exam yields scores which are normally distributed with a mean of 81 and a standard deviation of 5.5. if the civil service wishes to set a cut-off score on the exam so that 15% of the test takers fail the exam, what should the cut-off score be? remember to round your z-value to 2 decimal places.

The cutoff score should be 75.8 marks

Step-by-step explanation:

The cut-off should be set as the value corresponding to an area 15% in the normal distribution diagram.

Using the standard distribution tables we have value of standard normal deviate (Z) corresponding to area of 15% = -1.04

Thus we have

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."