scores on a University exam are normally distributed with a mean of 68 and a standard deviation of 9. use the 68-95-99.7 rule to answer the following questions
1) what proportions of students score between a 59 to 77?
2) what proportions of students score between a 50 to 86?
3) what proportions of students score above a 86?
4) what proportions of students score below a 41?
5) what proportions of students score above a 59?
6) what must a student score on the exam to be in the bottom 2.5% of scores on the exam?
Do you know the 68-95-99.7 rule? Approximately 68% of scores in normal distribution are within one standard deviation (34% on each side of the mean), 95% within 2 SD, and 99.7% within 3 SD.
Z = (score-mean)/SD
You can also find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores.
To answer these questions, we will use the 68-95-99.7 rule, which states that for a normal distribution:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
Now we can proceed to answer each question individually:
1) To find the proportion of students who score between 59 and 77, we need to find the area under the curve between these two scores. We calculate the z-scores for the lower and upper limits:
Lower z-score = (59 - 68) / 9 = -1
Upper z-score = (77 - 68) / 9 = 1
By referencing the z-table, we find that the area to the left of a z-score of -1 is approximately 0.1587, and the area to the left of a z-score of 1 is approximately 0.8413.
The proportion of students scoring between 59 and 77 is approximately 0.8413 - 0.1587 = 0.6826, or 68.26%.
2) To find the proportion of students who score between 50 and 86, we use the same process:
Lower z-score = (50 - 68) / 9 = -2
Upper z-score = (86 - 68) / 9 = 2
Using the z-table, we find that the area to the left of a z-score of -2 is approximately 0.0228, and the area to the left of a z-score of 2 is approximately 0.9772.
The proportion of students scoring between 50 and 86 is approximately 0.9772 - 0.0228 = 0.9544, or 95.44%.
3) To find the proportion of students who score above 86, we need to find the area under the curve to the right of this score:
z-score = (86 - 68) / 9 = 2
Using the z-table, we find that the area to the left of a z-score of 2 is approximately 0.9772.
The proportion of students scoring above 86 is approximately 1 - 0.9772 = 0.0228, or 2.28%.
4) To find the proportion of students who score below 41, we need to find the area under the curve to the left of this score:
z-score = (41 - 68) / 9 = -3
Using the z-table, we find that the area to the left of a z-score of -3 is approximately 0.0013.
The proportion of students scoring below 41 is approximately 0.0013, or 0.13%.
5) To find the proportion of students who score above 59, we need to find the area under the curve to the right of this score:
z-score = (59 - 68) / 9 = -1
Using the z-table, we find that the area to the left of a z-score of -1 is approximately 0.1587.
The proportion of students scoring above 59 is approximately 1 - 0.1587 = 0.8413, or 84.13%.
6) To find the score a student must achieve to be in the bottom 2.5% of scores, we need to find the z-score that corresponds to this proportion:
z-score = invNorm(0.025) ≈ -1.96
Now we can solve for the score using the formula:
x = z * σ + μ
x = -1.96 * 9 + 68
x ≈ 50.44
The student must score approximately 50.44 to be in the bottom 2.5% of scores on the exam.
To answer these questions using the 68-95-99.7 rule, you need to understand how to use the standard deviation and the mean to calculate proportions associated with different intervals.
In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations.
Now, let's calculate the proportions for each question:
1) To find the proportion of students who score between 59 and 77, we first need to convert these scores into z-scores. The z-score formula is (x - mean) / standard deviation. For x = 59, the z-score is (59 - 68) / 9 = -1. For x = 77, the z-score is (77 - 68) / 9 = 1. Therefore, we're looking for the proportion between z = -1 and z = 1. According to the 68-95-99.7 rule, approximately 68% of the data falls within one standard deviation of the mean. Hence, approximately 68% of the students score between 59 and 77.
2) Similar to the first question, we need to calculate the z-scores for x = 50 and x = 86. The z-score for x = 50 is (50 - 68) / 9 = -2, and for x = 86, it's (86 - 68) / 9 = 2. We're interested in the proportion between z = -2 and z = 2. According to the 68-95-99.7 rule, approximately 95% of the data falls within two standard deviations of the mean. Therefore, approximately 95% of students score between 50 and 86.
3) To find the proportion of students who score above 86, we want to know the proportion of students beyond z = 2. According to the 68-95-99.7 rule, only approximately 2.5% of the data falls beyond two standard deviations of the mean in the right tail. Hence, approximately 2.5% of students score above 86.
4) Similarly, to find the proportion of students who score below 41, we need to find the proportion beyond z = -3. According to the 68-95-99.7 rule, approximately 0.15% of the data falls beyond three standard deviations of the mean in the left tail. Therefore, approximately 0.15% of students score below 41.
5) To find the proportion of students who score above 59, we calculate the proportion beyond z = -1. According to the 68-95-99.7 rule, approximately 16% of the data falls beyond one standard deviation to the right. Therefore, approximately 16% of students score above 59.
6) To find the score a student must achieve to be in the bottom 2.5%, we need to calculate the z-score associated with that proportion. According to the 68-95-99.7 rule, approximately 2.5% of the data falls beyond two standard deviations to the left. Therefore, we need to find the z-score that corresponds to this proportion. Using a z-score table or calculator, the z-score at the 2.5% lower tail is approximately -1.96. Now, let's use the z-score formula to find the score: x = (-1.96 * 9) + 68 = 50.24. Hence, a student must score around 50.24 to be in the bottom 2.5% of scores on the exam.
Remember, these proportions are approximate estimates based on the assumptions of a normal distribution. In practice, the actual proportions may differ slightly.