The mean GPA of student in a course at UCDevis is 3.2 with a standard deviation of 0.3. What percent of student in a course have a GPA between 2.9 and 3.8?

81.86%

Step-by-step explanation:

We are given that the mean GPA of students in a course at UC Davis is 3.2 with a standard deviation of 0.3.

Assuming that the data follows normal distribution.

Let X = GPA of students in a course at UC Davis

So, X ~ Normal()

The z score probability distribution for normal distribution is given by;

                              Z  =   ~ N(0,1)

where,  = population mean GPA = 3.2

           = standard deviation = 0.3

Now, the probability that the students in the course have a GPA between 2.9 and 3.8 is given by = P(2.9 < X < 3.8)

       P(2.9 < X < 3.8) = P(X < 3.8) - P(X  2.9)

       P(X < 3.8) = P(  <  ) = P(Z < 2) = 0.97725

       P(X  2.9) = P(    ) = P(Z  -1) = 1 - P(Z < 1)

                                                        = 1 - 0.84134 = 0.15866

The above probability is calculated by looking at the value of x = 2 and x = 1 in the z table which has an area of 0.97725 and 0.84134 respectively.

Therefore, P(2.9 < X < 3.8) = 0.97725 - 0.15866 = 0.8186

Hence, 81.86% of students in the course have a GPA between 2.9 and 3.8.

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step-by-step explanation:

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