Answer :
Using the normal distribution, it is found that:
a) The interval that will contain most of the GPA's is (1.79, 4.43).
b) The z-score associated with this grade is of -2.82 < -2, thus, it is unlikely that this student studied 3 hours per day.
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Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
- If |Z| > 2, the measure X is considered to be unlikely.
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- Mean of 3.11 means that [tex]\mu = 3.11[/tex]
- Standard deviation of 0.66 means that [tex]\sigma = 0.66[/tex]
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Item a:
- This interval is composed by measures within 2 standard deviations of the mean, thus:
[tex]3.11 - 2(0.66) = 1.79[/tex]
[tex]3.11 + 2(0.66) = 4.43[/tex]
The interval that will contain most of the GPA's is (1.79, 4.43).
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Item b:
We have to find the z-score when [tex]X = 1.25[/tex], thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1.25 - 3.11}{0.66}[/tex]
[tex]Z = -2.82[/tex]
The z-score associated with this grade is of -2.82 < -2, thus, it is unlikely that this student studied 3 hours per day.
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