Solution: We are given that the grade point average of undergraduate students at one university have a normal distribution, with mean, [tex]\mu=2.52[/tex] standard deviation,[tex]\sigma=0.42[/tex]
We have to find the percentage of the students whose grade point averages are no more than 3.36, i.e., P(x<3.36).
We need to first find the z-score.
[tex]z=\frac{x-\mu}{\sigma}[/tex]
 [tex]=\frac{3.36-2.52}{0.42}[/tex]
 [tex]=\frac{0.84}{0.42}=2[/tex]
Now we have to find [tex]P(z<2)[/tex]. Using the empirical rule, we know that 97.5% data lies below 2 standard deviations above mean.
Therefore, using the empirical rule, 97.5% of the students have grade point averages that are no more than 3.36.
The percentage of the students have grade point averages that are no more than is 97.5%
The empirical rule states that for a normal distribution, 68% of the distribution are within one standard deviation from the mean, 95% are within two standard deviation from the mean and 99.7% are within three standard deviations from the mean.
Given that:
Mean (μ) = 2.52, Standard deviation (σ) = 0.42
68% are within one standard deviation = μ ± σ = 2.52 ± 0.42 = (2.12, 2.94)
95% are within two standard deviation = μ ± 2σ = 2.52 ± 2*0.42 = (1.68, 3.36)
The percentage of the students have grade point averages that are no more than 3.36 = 95% + (100% - 95%)/2 = 97.5%
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