X is a Gaussian random variable with mean 3 and standard deviation 2. Find (i) P[X2>2X] (8 pts), and (ii) expected value of [2X2-3X-5].

(i) Complete the square:

P[X ² > 2 X] = P[X ² - 2 X > 0]

P[X ² > 2 X] = P[X ² - 2 X + 1 > 1]

P[X ² > 2 X] = P[(X - 1)² > 1]

P[X ² > 2 X] = P[|X - 1| > 1]

P[X ² > 2 X] = P[X - 1 > 1] + P[-(X - 1) > 1]

P[X ² > 2 X] = P[X > 2] + P[X < 0]

P[X ² > 2 X] ≈ 0.7583

(ii) By definition of expectation, we have

E[2 X ² - 3 X - 5] = E[2 X ²] + E[-3 X] + E[-5]

E[2 X ² - 3 X - 5] = 2 E[X ²] - 3 E[X] - 5

Recall the definition of variance,

V[X] = E[(X - E[X])²] = E[X ²] - E[X]²

and that variance is the square of the standard deviation. So we have

E[2 X ² - 3 X - 5] = 2 (V[X] + E[X]²) - 3 E[X] - 5

E[2 X ² - 3 X - 5] = 2 (2² + 3²) - 3² - 5

E[2 X ² - 3 X - 5] = 12

2 x 503

step-by-step explanation:

its 1006

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.