Answer to Question #212075 in Statistics and Probability for Millicent

Question

5.1 Vehicles pass through a junction on a busy road at an average rate of 60 per hour.

5.1.1 Find the probability that none passes in a given 30-minute interval. (4)

5.1.2 What is the expected number passing in two minutes? (2)

5.1.3 Find the probability that this expected number (computed in 5.1.2) actually pass through in

a given five-minute period. (4)

5.2 Experience has shown that 80/200 of all CDs produced by a certain machine are defective. If a

 quality control technician randomly tests twenty CDs, compute each of the following

probabilities:

5.2.1 P (exactly one is defective).

5.2.2 P (at least one CD is defective).

5.2.3 P (no more than two are defective).

5.2.4 Find the mean, variance and standard deviation of the distribution.

(3) (4) (5)

Answer 1

5.1

$$\lambda=60(0.5)=30$$

5.1.1

$$P(X=0)=\dfrac{e^{-30}\cdot30^0}{0!}=e^{-30}\approx10^{-13}\approx 0$$

5.1.2

$$E(X)=60(\dfrac{2}{60})=2$$

5.1.3

$$\lambda=60(\dfrac{5}{60})=5$$

$$P(X=2)=\dfrac{e^{-5}\cdot5^2}{2!}=12.5e^{-5}\approx0.084224$$

5.2

$p=0.4, q=1-p=1-0.4=0.6, n=12$

5.2.1

$$P(X=1)=\dbinom{12}{1}(0.4)^1(0.6)^{12-1}$$

$$=0.01741425869$$

5.2.2

$$P(X\geq1)=1-P(X=0)$$

$$=1-\dbinom{12}{0}(0.4)^0(0.6)^{12-0}=$$

$$=0.99782321767$$

5.2.3

$$P(X\leq2)=P(X=0)+P(X=1)+P(X=2)$$

$$=\dbinom{12}{0}(0.4)^0(0.6)^{12-0}+\dbinom{12}{1}(0.4)^1(0.6)^{12-1}$$

$$+\dbinom{12}{2}(0.4)^2(0.6)^{12-2}=$$

$$=0.08344332288$$

5.2.4

$$\mu=np=12(0.4)=4.8$$

$$\sigma^2=npq=12(0.4)(0.6)=2.88$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{2.88}\approx1.697$$