An industrial psychologist has determined that the average percent score for an employee on a test of the employee's knowledge of the company's product is given by $P = \frac { 100 } { 1 + 24 e ^ { - 0.15 t } }$ where t is the number of weeks on the job and P is the percent score. Estimate (to the nearest week) the expected number of weeks of employment that are necessary for an employee to earn a 85% score on the test.
A)38 weeks
B)37 weeks
C)42 weeks
D)60 weeks
E)33 weeks

Question

Quizplus · Accepted Answer

To find the number of weeks (t) necessary for an employee to earn an 85% score, we set P = 85 and solve for t in the equation $P = \frac { 100 } { 1 + 24 e ^ { - 0.15 t } }$. Rearranging the equation to solve for t gives:$$85 = \frac { 100 } { 1 + 24 e ^ { - 0.15 t } }$$$$1 + 24 e ^ { - 0.15 t } = \frac { 100 } { 85 }$$$$24 e ^ { - 0.15 t } = \frac { 100 } { 85 } - 1$$$$e ^ { - 0.15 t } = \frac { 15 } { 85 \times 24 }$$$$e ^ { - 0.15 t } = \frac { 1 } { 136 }$$Taking the natural logarithm of both sides:$$-0.15 t = \ln \left( \frac { 1 } { 136 } \right)$$$$t = \frac { \ln \left( \frac { 1 } { 136 } \right) } { -0.15 }$$$$t \approx 33$$Therefore, it takes approximately 33 weeks for an employee to earn an 85% score on the test.

An Industrial Psychologist Has Determined That the Average Percent Score P=1001+24e−0.15tP = \frac { 100 } { 1 + 24 e ^ { - 0.15 t } }P=1+24e−0.15t100​

An Industrial Psychologist Has Determined That the Average Percent Score $P = \frac { 100 } { 1 + 24 e ^ { - 0.15 t } }$