Question 1

A call centre receives 30 calls per minute on average. What is the probability that it will receive exactly 40 calls in a minute (3 decimal places)?

Accepted Answer

The scenario can be modeled using a Poisson distribution, where the average rate ($\lambda$) is 30 calls per minute. The probability of observing $k$ events (in this case, calls) in a fixed interval is given by the formula: $$P(k; \lambda) = \frac{e^{-\lambda} \lambda^k}{k!}$$ For exactly 40 calls, $k = 40$ and $\lambda = 30$, so: $$P(40; 30) = \frac{e^{-30} \cdot 30^{40}}{40!}$$ Calculating this gives a probability of approximately 0.014, hence the correct answer is B.

Question 2

In a binomial trial of p = 0.35 and n = 10 what are the chances of x < 2?

Accepted Answer

D

Question 3

Which of the following options does NOT describe a binomial situation?

Accepted Answer

Option B does not describe a binomial situation because it involves continuous data (value of sales), whereas a binomial situation requires outcomes that can be classified into two categories (e.g., success/failure, yes/no, defective/good).

Question 4

In a binomial situation with p = 0.24 and n = 12 what is the probability of x = 3 (2 decimal places)?

Accepted Answer

The answer of In a binomial situation with p =...

Question 5

Which of the following options is NOT an example of a poisson situation?

Accepted Answer

The Poisson distribution is typically used to model the number of times an event happens in a fixed interval of time or space. Options A, B, and C fit this description as they deal with counts of events (defects, complaints, breakdowns) in a given time frame. Option D, however, refers to the outcomes of calls (successful or unsuccessful), which is more appropriately modeled by a binomial distribution, where each call has two possible outcomes (success or failure), rather than the count of events in a fixed interval.

Question 6

Use a poisson approximation to the binomial to work out the probability that there will be less than two faulty electric motors in a box of 100 with a defect rate of 0.5% (as a percentage, 1 decimal place)

Accepted Answer

The Poisson approximation is suitable here because the number of trials (n=100) is large and the probability of defect (p=0.005) is small. The mean (λ) for the Poisson distribution is np = 100 * 0.005 = 0.5. The probability of having less than two faulty motors is P(X<2) = P(X=0) + P(X=1) = e^(-0.5) * (0.5^0/0!) + e^(-0.5) * (0.5^1/1!) = 0.6065 + 0.3033 = 0.9098, which is approximately 90.98% or 4.0% when expressed as a percentage to one decimal place.

Question 7

A Swedish timber mill produces planks of wood. These are tested and rejected if they are too thick or too thin. If samples of 200 planks are tested and the chance of any single plank being rejected is 2%, what is the mean number of planks in each sample that will be rejected?

Accepted Answer

The answer of A Swedish timber mill produces planks of...

Question 8

In a poisson situation where m = 0.85 find the probability that x = 2 (as a percentage, 1 decimal place)

Accepted Answer

The probability of x = 2 in a Poisson distribution with m = 0.85 is calculated using the formula P(x) = (e^(-m) * m^x) / x!. Substituting the values, P(2) = (e^(-0.85) * 0.85^2) / 2! = 0.154, which is 15.4% when expressed as a percentage.

Question 9

On average a taxi firm has 1.5 breakdowns every 2 weeks. Find the probability that there will be 1 or 2 breakdowns in a 4 week period.

Accepted Answer

The answer of On average a taxi firm has 1.5...

Question 10

A IT distribution company in Hong Kong gives a bonus to salespeople that generate more than HK$180,000 in a single day. On average this happens 5% of the time. If the company employs 20 salespeople, what is the probability of paying exactly 1 bonus on any particular day (give as a percentage, 0 decimal places)?

Accepted Answer

The probability of exactly 1 salesperson out of 20 generating more than HK$180,000 in a single day, and thus earning a bonus, can be calculated using the binomial probability formula. The formula for exactly k successes (in this case, 1 bonus) out of n trials (in this case, 20 salespeople) is given by:$$ P(k; n, p) = \binom{n}{k} p^k (1-p)^{n-k} $$where $\binom{n}{k}$ is the binomial coefficient, $p$ is the probability of success on a single trial (5% or 0.05), and $1-p$ is the probability of failure on a single trial.For exactly 1 bonus:$$ P(1; 20, 0.05) = \binom{20}{1} (0.05)^1 (0.95)^{19} $$$$ P(1; 20, 0.05) = 20 \times 0.05 \times (0.95)^{19} $$$$ P(1; 20, 0.05) \approx 0.377 $$$$ P(1; 20, 0.05) \approx 38\% $$Therefore, the probability of paying exactly 1 bonus on any particular day is approximately 38%.

Quiz 33: Binomial and Poisson Distributions

A call centre receives 30 calls per minute on average. What is the probability that it will receive exactly 40 calls in a minute (3 decimal places) ?