Question

Scores on exam are normally distributed with a mean of 45 and standard deviation of 8

Answer 1

1.       First, we solve the z-score of 52.6

                                                z-score = (52.6 – 45)/81 = 0.0938

                                            percentile value = 53.7

Therefore, the probability is 100-53.7 = 46.3%

 

2.       We solve again for the z-score of 43

                                                z-score = (43 – 45)/81 = -0.0247

                                                percentile value = 49%

Thus, 49% of students scored 49%

 

3.       Z-score of 49,

z-score = (49-45)/81 = 0.04938

 

4.       If z-score is 1.5,

1.5 = (mean – 45)/81 = 166.5