Question

The business school at Wassamatta U. receives a large number of applicants to its undergraduate honors program in economics. This Excel file Honors Program GPA shows the GPA's of a sample of 260 applicants that have applied to the program in the last several years. From the data the mean GPA is

xˉ =3.13 with standard deviation s=0.395. Question 1. If a Normal model N(3.13, 0.395) is an appropriate model for the GPA's of all applicants to the undergradua honors program in economics, what is the probability that an applicant's GPA is 3.25 or greater? Use 3 decimal places

Answer 1

The probability that an applicant's GPA is 3.25 or greater, assuming a Normal model N(3.13, 0.395), is approximately 0.382.

Step-by-step explanation:

In order to find the probability that an applicant's GPA is 3.25 or greater, we can use the standard normal distribution and z-scores. The formula for the z-score is given by
`\( z = \frac{(X - \bar{x})}{s} \),` where X is the value we're interested in,
`\( \bar{x} \)` is the mean, and s is the standard deviation. In this case, for a GPA of 3.25:

`\[ z = ((3.25 - 3.13))/(0.395) \]`

Calculating this, we find
`\( z \approx 0.304 \).`To find the probability of a z-score being less than or equal to 0.304, we can consult a standard normal distribution table or use a calculator. The cumulative probability for a z-score of 0.304 is approximately 0.617. Since we want the probability of the GPA being 3.25 or greater, we subtract this value from 1:

`\[ P(X \geq 3.25) = 1 - P(X \leq 3.25) \]`

`\[ \approx 1 - 0.617 \]`

`\[ \approx 0.382 \]`

Therefore, the probability that an applicant's GPA is 3.25 or greater is approximately 0.382. This implies that around 38.2% of applicants have a GPA of 3.25 or higher, according to the provided Normal model.