Answer:
a) The value of c that will make the density function a probability density function = 0.0430
b) P(X=2) = 0.387
P(X > 2) = 0.35475 = 0.355 to 3 s.f
c) The probability that the first random variable/trial to be greater than 2 is on the 10 trial (first 9 trials are less than 2 and the 10 trial is greater than 2) = 0.0000018125 = (1.8125 × 10⁻⁶)
The probability of getting a value greater than 2 on the 10th trial and getting a Value NOT greater than 2 on first 9 trials = 0.006878
d) 0.9806
Explanation:
Continuous random variable X with density function f(x) = c (1 + x³) where c = constant
with a support SX = [0,3]
a) Value of c that will make the density function a valid probability density function.
A valid probability density function sums up to give 1 over the interval of its sample space. That is, the sum of the probabilities over its sample space equals 1.
P(U) = ∫³₀ f(x) dx = 1
∫³₀ c (1 + x³) dx = 1
c ∫³₀ (1 + x³) dx = 1
c [x + (x⁴/4)]³₀ = 1
c [3 + (3⁴/4)] = 1
23.25c = 1
c = (1/23.25)
c = 0.0430
The value of c that will make the density function a probability density function = 0.0430
b) Probability that X = 2
f(x) = 0.043 (1 + x³)
P(X=2) = f(2) = 0.043 (1 + 2³) = 0.043 (1 + 8)
P(X=2) = 0.043 × 9 = 0.387
Probability that X is greater than 2 = P(X > 2)
P(2 ≤ X ≤ 3) = ∫³₂ f(x) dx
P(2 ≤ X ≤ 3) = ∫³₂ 0.043 (1 + x³) dx
= 0.043 ∫³₂ (1 + x³) dx
= 0.043 [x + (x⁴/4)]³₂
= 0.043 {[3 + (3⁴/4)] - [2 + (2⁴/4)]}
= 0.043 {23.25 - 6}
= 0.043 × 17.25
P(2 ≤ X ≤ 3) = P(X ≥ 2) = 0.74175
P(X > 2) = P(X ≥ 2) - P(X=2)
= 0.74175 - 0.387 = 0.35475
c) the probability that the first random variable/trial to be greater than 2 is on the 10 trial (first 9 trials are less than 2 and the 10 trial is greater than 2)
P(X>2) = 0.35475
P(X<2) = 1 - P(X ≥ 2) = 1 - 0.74175 = 0.25825
The required probability = [P(X<2)]⁹ × [P(X>2)] = (0.25825)⁹ × (0.35475) = 0.0000018125 = (1.8125 × 10⁻⁶)
Although, this probability could also be interpreted as the probability of getting a Value greater than 2 on the 10th trial and getting a value NOT greater than 2 on first 9 trials.
P(X>2) = 0.35475
Probability of X not greater than 2 = P(X<2) + P(X=2) = 1 - 0.35475 = 0.64525
Required probability = [P(X≤2)]⁹ × P(X>2) = (0.64525)⁹ × 0.35475 = 0.006878
d) Probability that it will take less than 10 random variables/trials before we see a trial that is greater than 2?
This is a sum of probabilities from getting a trial greater than 2 on the first attempt to getting it on the 9th attempt (less than 10 trials)
P(X>2) = 0.35475
Probability of X not greater than 2 = P(X<2) + P(X=2) = 1 - 0.35475 = 0.64525
Required probability = [0.35475 + (0.64525)(0.35475) + (0.64525)²(0.35475) + (0.64525)³(0.35475) + (0.64525)⁴(0.35475) + (0.64525)⁵(0.35475) + (0.64525)⁶(0.35475) + (0.64525)⁷(0.35475) + (0.64525)⁸(0.35475)] = 0.980611147 = 0.9806
Hope this Helps!!!