The amount of time (in minutes) that a commuter train is late is a continuous random variable with the probability density function listed below. Find the mean and variance of the amount of time in minutes the train is late. (Note: A negative time value means that the train is early).

Mean waiting time = E(X) = 0 minute

Variance = Var(X) = 140

Step-by-step explanation:

f(x) = –3(25 – x²)/500 for –10 < x < +10

f(x) = (-75 + 3x²)/500 for -10 < x < 10

f(x) = (0.006x² - 0.15) for -10 < x < 10

a) The mean of a Probability distribution is given by its expected value

Mean = E(X) = Σ xᵢpᵢ

xᵢ = each variable In the sample space

pᵢ = probability of each variable In the sample space.

In integral terms,

E(X) = ∫ xf(x) dx (evaluating the integral all over the sample space; that is, for - 10 to 10)

E(X) = ∫¹⁰₋₁₀ xf(x) dx

E(X) = ∫¹⁰₋₁₀ x[0.006x² - 0.15] dx

E(X) = ∫¹⁰₋₁₀ [0.006x³ - 0.15x] dx

E(X) = [0.0015x⁴ - 0.075x²]¹⁰₋₁₀

E(X) = [0.0015(10⁴) - 0.075(10²)] - [0.0015(-10⁴) - 0.075(-10²)]

E(X) = [15 - 7.5] - [15 - 7.5] = 0 minutes

b) Variance is given as

Variance = Var(X) = Σx²p − μ²

where μ = mean = E(X)

In integral terms,

Var(X) = [∫¹⁰₋₁₀ x²f(x) dx] - μ²

∫¹⁰₋₁₀ x²f(x) dx = ∫¹⁰₋₁₀ x²(0.006x² - 0.15) dx

= ∫¹⁰₋₁₀ (0.006x⁴ - 0.15x²) dx

= [0.0012x⁵ - 0.05x³]¹⁰₋₁₀

= [0.0012(10⁵) - 0.05(10³)] - [0.0012(-10⁵) - 0.05(-10³)]

= [120 - 50] - [-120 + 50]

= 70 + 70 = 140

Var(X) = [∫¹⁰₋₁₀ x²f(x) dx] - μ²

Var(X) = 140 - 0² = 140.

Hope this Helps!!!

the answer is: -22 also a

hope i

-lavira: )