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The American Economic Review (December 2008) published a study on the factors (other than credit constraints) that would influence the college dropout decision. One factor of interest was expected GPA for a college student who studied 3 hours per day. In a representative sample of 307 college students, the researchers reported the following summary statistics for the expected GPA of those who studied 3 hours per day: x 3.11, s 66. a. Give an interval that will contain most (at least 75% to approximately 95%) of the 307 GPAs. b. If you observe a GPA of 1.25, is it likely that this college student studied 3 hours per day? Explain.

Sagot :

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.

A similar problem is given at https://brainly.com/question/16347292

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