Question

Binomial Distribution: 12:17

Notes:

the binomial distribution is a discrete probability distribution

the parameters of binomial distribution are p (probability of success in a single trial) and n (number of trials)

the mean of binomial distribution is np

the standard deviation of binomial distribution is sqrt(npq) q=1-p

1. In the production of bearings, it is found out that 3% of them are defective. In a randomly collected sample

of 10 bearings, what is the probability of

Given: p = 0.03, n = 10 q = 1-p = 0.97

a. 2 of them are defective = p(2) =

b. 3 of them are defective = p(3) =

2. If a production line makes 8 good parts out of 10 parts made, find the mean and standard deviation of good parts out of total 400 parts made?

Given: p = 8/10 = 0.8, n = 400 q = 1-p = 0.2

Mean number of good parts made out of 400 = np =

Standard deviation = sqrt(npq) =

3. Assuming the probability of making a perfect (good) part is ½, find the probability that a sample of 3 parts will have

Given: p = 0.5, q = 1-p = 0.5, n = 3

a. At least one good part

P(1 or 2 or 3 good parts) = 1 - P(x=0 bad parts) =

b. Two good parts

P(x=2 good parts) =

c. At most two good parts

P(0 good parts) or P(1 good part) or P(2 good parts) = P(x=0) + P(x=1) + P(x=2) =

d. At least two good parts

P(2 good parts) or P(3 good parts) = P(x=2) + P(x=3) = 3/8 + 1/8 = 4/8 =

4. The probability of making bad parts on a production line is 60%. If a sample of 5 parts are randomly

selected, how many of them have 3 or more good parts?

Given: p of making good parts = 0.4, p of making bad parts = q=1-p = 0.6

P(x=3 or x=4 or x=5) =

Poisson Distribution:

Notes:

it is a discrete probability distribution

the probability of success is very low

the number of trials (n) is very large

the trials are independent

the only parameter mean is λ = np is finite

the variance is λ

1. At a shopping center, the number of customers appear at the rate of 2 per minute. Find the probability of:\

Given: λ = 2/min

a. No customers appear P(x=0) =

b. Only one customer appears P(x=1) =

c. Only two customers appear P(x=2) =

d. At least two customers appear 1 – [P(x=0) + P(x=1)] =

e. At the most two customers appear P(x=0) + P(x=1) + P(x=2) =

f. Less than three customers appear P(x=0) + P(x=1) + P(x=2) =

2. The average number of car accidents on I-75 is 4 per day. Find the probability of

a. No accidents in a day P(x=0) =

b. At least 2 accidents in a day P(x≥2) = 1 – [P(x=0) + P(x=1)] =

c. At the most 3 accidents P(x≤3) = P(x=0) + P(x=1) + P(x=2) + P(x=3) =

Answer 1