Question

Problems 27: the number X is the number of car which passed through the

specific stop sign in 1 minutes :
X 0 1 2 3 4 5

P(x) a 0.30 0.05 0.03 0.02 0.01

a) find out the value of a. (1pt)

b). the probability that there are at least 3 cars passing through the stop sign.
(1pt)

c). the expected value of X. (1pt)

d). the variance of X. (1pt)

e) the standard deviation of X. (1pt)

Answer 1

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

Explanation:

Given data

X : 0 1 2 3 4 5

P(x) : a 0.30 0.05 0.03 0.02 0.01

a)

∑
`p(x_(i) )` = 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