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

1. what is the probability that the score of a student will be higher than 52.6?

2. what proportion of the students score below 43?

3. what is the z score that corresponds to the score 49?

what is the raw score than corresponds to z=1.5?

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