Question

At a community college it is known that the grade point average (GPA) of all students has a distribution which is both mound shaped and symmetric. The mean of the distribution is 2.50 with a standard deviation of 0.5. It has been decided by the Faculty Senate that all students with GPAs in the top 16% will graduate with honors. What is the minimum GPA that qualifies a student to graduate with honors? (Hint: you will need to use the Empirical Rule to answer this question) Select one:

Answer 1

The minimum GPA that qualifies a student to graduate with honors is 3.00.

Step-by-step explanation:

To determine the minimum GPA that qualifies a student to graduate with honors, we can use the Empirical Rule. According to the Empirical Rule, for a distribution that is mound shaped and symmetric, approximately 68% of the data falls within one standard deviation of the mean, approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations.

In this case, we want to find the minimum GPA that qualifies a student to be in the top 16%. Since the distribution is symmetric, the top 16% is equal to the upper 16% plus the lower 16%, which is 32%. So, we need to find the GPA that corresponds to the upper 32%.

Using the Empirical Rule, we know that the GPA that corresponds to one standard deviation above the mean is approximately the mean plus one standard deviation. So, the minimum GPA that qualifies a student to graduate with honors is 2.50 + 0.5 = 3.00.

Answer 2

3 GPA is the minimum GPA that qualifies a student to graduate with honors.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.50

Standard Deviation, σ = 0.5

We are given that the distribution of GPA score is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

We have to find minimum GPA in the top 16% students will graduate with honors.

Thus,

`P(z) = 1 - 0.16 = 0.84`

Calculating the corresponding value of z from the normal distribution table, we have,

z = 0.994458

`\displaystyle(x-\mu)/(\sigma) = 0.994458\\\\(x - 2.5)/(0.5) = 0.994458\\\\x = (0.994458* 0.5) + 2.5\\x = 2.997229 \approx 3`

Hence, 3 GPA is the minimum GPA that qualifies a student to graduate with honors.