Section A

1.) The weight X of a particular animal is normally distributed with μ= 200kg and σ= 15kg. An animal of this population is overweight if it has a weight greater than 230 kg

a) Find the probability than an animal is overweight.
P (X>230) = 0.02275

b) We select 2 animals from this population. Find the probabilities that

i) both animals are overweight
(0.02275)² = 0.000518

ii) only one animal is overweight.
0.02275)* (1-0.02275)* 2 = 0.0445

c) We select 7 animals of this population. Find the probability that exactly two of them are overweight.
Y follows B(n,p) with n = 7, p =0.02275
P(Y=2) = 0.00969

2.) Consider the independent events A and B. Given that P(B) = 2P(A) and P(A⋃B) = 0.52. Find P(B).

Let P(A)= x and P(B)= 2x
0.52 = x + 2x – 2x²
x= 0.2
P(B) = 0.4

3.) Let f(x) = x³ – 3x² + 2. Find the point of inflection of the graph of y = f(x).

To find the point of inflection, we need to find the critical points of the function and then determine the concavity. The critical points are found by setting the first derivative of the function to 0 and solving for x.
f'(x) = 3x² – 6x

Setting f'(x) = 0 and solving for x, we find two critical points:
x = 0 and x = 2

To determine the concavity, we need to find the second derivative of the function and evaluate it at the critical points.
f”(x) = 6x – 6
f”(0) = -6 and f”(2) = 6

At x = 0, f”(x) is negative, so the graph is concave down. At x = 2, f”(x) is positive, so the graph is concave up. Since the concavity changes at x = 2, this is the point of inflection. Therefore, the point of inflection of the graph of y = f(x) is (2, f(2)) = (2, -2).

4.) A farmer has 600 meters of fencing and wants to enclose a rectangular area for his sheep. He wants the enclosed area to be as large as possible and has a stream running along one side of the rectangle that will serve as the fourth side.

Formulate the problem as a calculus problem and find the maximum area.

We can formulate this problem as an optimization problem using calculus. Let x be the length of the rectangle and y be the width. We know that the total fencing available is 600 meters and that the rectangle has two sides of length x and two sides of length y.
Therefore, the expression for the total fencing used is 2x + 2y = 600.

The area of the rectangle is given by A = xy. We want to maximize this area subject to the constraint
2x + 2y = 600.

We can start by isolating y in the constraint equation: y = -x + 300

Now we can substitute this expression for y into the area expression and get A = x(-x + 300) = -x² + 300x
We can find the critical points by taking the derivative of the area expression with respect to x and setting it equal to 0:
A'(x) = -2x + 300 = 0
x = 150

Now we need to check if this critical point is a maximum or minimum. We can do this by taking the second derivative of the area expression and evaluating it at the critical point:
A”(x) = -2

Since the second derivative is negative, the critical point x = 150 is a maximum. We can substitute this value of x back into the constraint equation to find the corresponding value of y:
y = -x + 300 = -150 + 300 = 150

Therefore, the maximum area is 150*150 = 22500 square meters, and the dimensions of the rectangle are 150m by 150m.

5.) Consider the following data:

x	1	2	3	4
y	2	3	7	8

a) Find the mean and the variance for the values of x.
Mean= 2.5, Variance= 1.11803² = 1.25

b)Find the mean and the variance for the values of y.
Mean= 5, Variance= 2.545951² = 6.5

c)Find the correlation coefficient r.
0.965

d)Describe the relation between x and y.
Strong positive