Answer:
a) The value of a= 0.59
b) The probability that there are at-least 3 cars passing through the stop sign
P(x>3) = 0.03
c)
The Expected value of X = 0.62
d)
The variance of X is σ² = 0.9556
e)
The standard deviation of X
σ = 0.9775
Step-by-step explanation:
Given data
X : 0 1 2 3 4 5
P(x) : a 0.30 0.05 0.03 0.02 0.01
a)
∑ [tex]p(x_{i} )[/tex] = 1
a + 0.30+ 0.05+ 0.03+ 0.02+0.01 = 1
a + 0.41 = 1
a = 1 - 0.41
a = 0.59
b)
The probability that there are at-least 3 cars passing through the stop sign
P(x >3) = P( x=4) + P( x=5)
= 0.02 +0.01
= 0.03
c)
X : 0 1 2 3 4 5
P(x) : 0.59 0.30 0.05 0.03 0.02 0.01
The Expected value of X
E(X) = ∑ x P(X= x)
= 0 + 1 ×0.30 + 2×0.05 + 3×0.03 + 4×0.02 + 5×0.01
= 0.30 + 0.1 + 0.09 +0.08 +0.05
= 0.62
The Expected value of X
E(X) = 0.62
d)
The variance of the discrete distribution
σ² = ∑ x²p(x) -μ²
σ² = 0 + 1² ×0.30 + 2² ×0.05 + 3² ×0.03 + 4² ×0.02 + 5²× 0.01 - (0.62)²
= 1.34 - 0.3844
= 0.9556
σ² = √0.9556
e) The standard deviation of the discrete distribution
σ = 0.9775
