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

2 is the final ! : )

a

step-by-step explanation: