The mean GPA of students in a course at UCDavis is 3.2 with a standard deviation of 0.3.

What percent of students in the course have a GPA between 2.9 and 3.8?

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

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.

2.56 is the width of the garden

step-by-step explanation:

i believe the answer is c.