Answer :
Answer:
Using the 68-95-99.7 rule, approximately 84% of students scored 90 or less on the university exam.
Step-by-step explanation:
The 68-95-99.7 rule is a statistical rule that applies to a normal distribution. It states that:
- 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.
In this case, we are given that the scores on the university exam are Normally distributed with a mean (μ) of 86 and a standard deviation (σ) of 4.
To find the percentage of students who scored 90 or less, we need to calculate the z-score for 90:
\[Z = \frac{X - μ}{σ} = \frac{90 - 86}{4} = 1\]
A z-score of 1 corresponds to being one standard deviation above the mean. Since the mean is 86 and one standard deviation is 4, this means that 90 is one standard deviation above the mean.
Using the 68-95-99.7 rule, we know that approximately 68% of students scored within one standard deviation of the mean or 86 ± 4. Therefore, approximately 68% of students scored 90 or less.
Learn more about the normal distribution:
brainly.com/question/34741155
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