Suppose the scores of students on an exam are Normally distributed with a mean of 297 and a standard deviation of 66. Then approximately 99.7% of the exam scores lie between the numbers _____and _______such that the mean is halfway between these two integers.
To find the numbers between which approximately 99.7% of the exam scores lie, we can use the empirical rule.
According to the empirical rule, in a normal distribution:
- Approximately 68% of the data falls within 1 standard deviation of the mean.
- Approximately 95% of the data falls within 2 standard deviations of the mean.
- Approximately 99.7% of the data falls within 3 standard deviations of the mean.
In this case, the mean of the distribution is 297 and the standard deviation is 66.
So, within 3 standard deviations, we can calculate the values as follows:
Lower value: Mean - (3 * Standard Deviation)
= 297 - (3 * 66)
= 99
Upper value: Mean + (3 * Standard Deviation)
= 297 + (3 * 66)
= 495
Therefore, approximately 99.7% of the exam scores lie between the numbers 99 and 495, and the mean, 297, is halfway between these two integers.
To find the approximate range of exam scores that includes 99.7% of the scores, we can use the concept of z-scores.
Given that the scores follow a normal distribution with a mean of 297 and a standard deviation of 66, we can calculate the z-scores corresponding to the 99.7% interval.
To find the z-scores, we need to use the standard normal distribution table or a statistical calculator. The z-scores represent the number of standard deviations a particular value is from the mean.
For the lower bound, we need to find the z-score that corresponds to the percentile at which 0.15% (half of 0.3%) of the scores lie below it. Similarly, for the upper bound, we need to find the z-score that corresponds to the percentile at which 0.15% of the scores lie above it.
Since the standard normal distribution is symmetric, we can find these z-scores by subtracting and adding the desired percentage from 50%. This is because 50% of the scores will lie on either side of the mean.
Let's calculate the z-scores:
For the lower bound:
Lower z-score = 50% - (0.15%/2) = 0.5 - 0.0075 = 0.4925
For the upper bound:
Upper z-score = 50% + (0.15%/2) = 0.5 + 0.0075 = 0.5075
Next, we can convert these z-scores back into exam scores by using the z-score formula:
z = (X - mean) / standard deviation
For the lower bound, solving for X:
0.4925 = (X - 297) / 66
X - 297 = 0.4925 * 66
X - 297 = 32.49
X = 32.49 + 297
X ≈ 329.49
For the upper bound, solving for X:
0.5075 = (X - 297) / 66
X - 297 = 0.5075 * 66
X - 297 = 33.495
X = 33.495 + 297
X ≈ 330.495
Therefore, approximately 99.7% of the exam scores lie between roughly 329.49 and 330.495, with the mean (297) halfway between these two values.
363-231
you can play around with Z table stuff at
http://davidmlane.com/hyperstat/z_table.html
It has very nice input forms so you can specify tails or middle regions.