Solve the problem.
-Scores on a test have a mean of 70 and $\mathrm { Q } _ { 3 }$ is 83. The scores have a distribution that is approximately normal. Find P90. (You will need to first find the standard deviation.)
A) 93.7
B) 95.6
C) 94.8
D) 92.9

Question

Quizplus · Accepted Answer

The z-score for Q3 is (83-70)/$\sigma$=1.3. Using a standard normal distribution table, we find that the z-score for P90 is 1.28. Therefore, P90 = 70 + 1.28$\sigma$ = 94.8.

Solve the Problem Q3\mathrm { Q } _ { 3 }Q3​

Solve the Problem $\mathrm { Q } _ { 3 }$