Final answer:
a) Using the 68-95-99.7 rule, we expect around 84.13% of scores to fall above 45. b) We expect approximately 168 scores to fall above 45. c) The z-score corresponding to an exam score of 45 is -2.
Step-by-step explanation:
a) To determine the percent of scores expected to fall above 45, we need to find the area under the normal curve to the right of the z-score corresponding to 45. Using the 68-95-99.7 rule, we know that approximately 68% of the scores fall within one standard deviation of the mean, so the area to the left of the z-score of -1 is 0.1587. Therefore, the area to the right of the z-score of -1 (corresponding to 45) is 1 - 0.1587 = 0.8413, or 84.13%.
b) To find the number of scores expected to fall above 45, we multiply the percentage from part a) by the total number of students in the class. So, the number of scores expected to fall above 45 is 0.8413 * 200 = 168.26, or approximately 168.
c) The z-score corresponding to the exam score of 45 can be found using the formula z = (x - mean) / standard deviation. Plugging in the values, we get z = (45 - 65) / 10 = -2, so the z-score is -2.
d) To find the percent of scores expected to fall between 65 and 75, we need to find the area under the normal curve between the corresponding z-scores. Using the z-table, we can find that the area to the left of a z-score of 1 (corresponding to 75) is 0.8413, and the area to the left of a z-score of 0 (corresponding to 65) is 0.5. So, the area between these two z-scores is 0.8413 - 0.5 = 0.3413, or 34.13%.
e) To calculate the exam score associated with a z-score of 1.2, we use the formula x = mean + (z * standard deviation). Plugging in the values, we get x = 65 + (1.2 * 10) = 77.
f) To calculate the z-score associated with an exam score of 81, we use the formula z = (x - mean) / standard deviation. Plugging in the values, we get z = (81 - 65) / 10 = 1.6.
the complete Question is given below:
Suppose that the distribution of exam scores in a class of 200 student is a normal distribution. Also suppose that the distribution’s mean is 65 points, and its standard deviation is 10.
a) Use the 68-95-99.7 rule to determine what percent of the scores do we expect to fall above 45 _______%
b) How many scores do we expect to fall above 45 ? _______
c) Which z-score corresponds to the exam score from port a and b? _______
d) What percent of scores do we expect io tall between 65 and 75 ? _______%
e) Calculate the Exam score associated with the z-score z=12. _______
f) Calculate the z-score associated with the Exam score 81 . Round all answers to the second decimal place as needed