Question 1

Consider the following to answer the question(s) below:
A company's manufacturing process uses 500 litres of water at a time. A "scrubbing" machine then removes most of a chemical pollutant before pumping the water into a nearby lake. To meet federal regulations the treated water must not contain more than 80 parts per million (ppm) of the chemical. Because there is a fine charged if regulations are not met, the company sets the machine to attain an average of 75 ppm in the treated water. The machine's output can be described by a Normal model with a standard deviation 4.2 ppm.
-The company's lawyers insist that not more than 2% of the treated water should be over the limit. In order to achieve this, to what mean should the company set the scrubbing machine? Assume the standard deviation does not change.
A) 75 ppm
B) 78.626 ppm
C) 71.373 ppm
D) 80 ppm
E) 60 ppm

Accepted Answer

Let X be the amount of chemical in the treated water. We want to find the mean of X such that P(X > 80) is at most 0.02.

We have P(X > 80) = P((X - 75)/4.2 > (80 - 75)/4.2) = P(Z > 1.19), where Z is the standard normal random variable.

Using a standard normal table or calculator, we find that P(Z > 1.19) is approximately 0.1179.

To ensure that P(X > 80) is at most 0.02, we need P(Z > k) to be 0.02, where k is the value such that P(Z > k) = 0.02.

From a standard normal table or calculator, we find that k is approximately 2.05.

Therefore, we have (80 - 75)/4.2 = 2.05, which gives a mean of X equal to 71.373 ppm.

Thus, the best choice is C.

Question 2

Consider the following to answer the question(s) below:
A company believes that there will be a 50% chance of making a profit of $1,000, a 30% chance of making a profit of $1500 and a 20% chance of making a profit of $2,000.
-Find the standard deviation for the company's profit?
A) 390.51
B) 152,498
C) 350
D) 18.71
E) 1350

Accepted Answer

To find the standard deviation of the profits, we need to first find the expected value (mean) of the profits.

Expected value = (0.5 x $1000) + (0.3 x $1500) + (0.2 x $2000) = $1350

Now we will find the variance:

Variance = (0.5 x ($1000-$1350)^2) + (0.3 x ($1500-$1350)^2) + (0.2 x ($2000-$1350)^2) 
         = $390,510

Finally, we take the square root of the variance to get the standard deviation:

Standard deviation = √($390,510) = $390.51

Therefore, the answer is A) $390.51.

Question 3

The owner of a pet store is trying to decide whether to discontinue selling specialty clothes for pets. She suspects that only 4% of the customers buy specialty clothes for their pets and thinks that she might be able to replace the clothes with more profitable items. Before making a final decision, she decides to keep track of the total number of customers for a day and whether they purchase specialty clothes.
a. The owner had 275 customers that day. Assuming this was a typical day for her store, what would be the mean and standard deviation of the number of customers who buy specialty clothes for their pet each day?
b. Surprised by the high number of customers who purchased specialty pet clothing that day, the owner decided that her 4% estimate must have been too low. How many clothing sales would it have taken to convince you? Justify your answer.

Accepted Answer

a. Using the Binomial model, Mean: μ = np = (275)(0.04) = 11
Standard deviation: σ =  =  = 3.25
b. Since np = 11 and nq = 264, we expect at least 10 successes and at least 10 failures. The sample size is large enough to apply a Normal model. It would be unusual to see the number of customers who purchased specialty pet clothing more than 2 or 3 standard deviations above the mean. Since the standard deviation = 3.25, it would be unusual to see more than 18 (11 + 2(3.25) = 17.5) customers who purchased specialty clothing. So, I would conclude that her 4% estimate must have been too low if more than 18 customers purchased specialty clothing for their pet.

Question 4

Consider the following to answer the question(s) below:
A courier service claims that no more than 5 % of all of its deliveries arrive late. Assuming deliveries are independent, a sample of 10 deliveries is randomly selected.
-What is the probability that more than 2 of the sample deliveries arrive late?
A) 0.0116
B) 0.0105
C) 0.0862
D) 0.0746
E) 0.0439

Accepted Answer

The probability of a delivery arriving late is 0.05. The probability of a delivery arriving on time is 0.95. The probability of more than 2 deliveries arriving late is the complement of the probability of 0, 1, or 2 deliveries arriving late. The probability of 0 deliveries arriving late is 0.95^10 = 0.5987. The probability of 1 delivery arriving late is 10 * 0.05 * 0.95^9 = 0.2843. The probability of 2 deliveries arriving late is 45 * 0.05^2 * 0.95^8 = 0.1161. The probability of more than 2 deliveries arriving late is 1 - 0.5987 - 0.2843 - 0.1161 = 0.0009.

Question 5

Consider the following to answer the question(s) below:
The number of clients arriving at a bank machine is Poisson distributed with an average of 2 per minute. For a 5 minute period,
-What is the probability that no more than 2 customers will arrive in a 5 minute period?
A) 0.0005
B) 0.1246
C) 0.0028
D) 0.6767
E) The answer cannot be determined with the given information.

Accepted Answer

The Poisson distribution is used when we want to know the probability of a certain number of events occurring in a fixed time interval when the events are occurring randomly and independently. In this case, the average rate of arrival is given as 2 customers per minute. Thus, the parameter λ for a 5-minute period will be 2*5=10. We need to find the probability that no more than 2 customers will arrive in a 5-minute period which translates to finding P(X ≤ 2) where X is a Poisson variable with mean λ =10. Using the Poisson distribution formula, we get:

P(X ≤ 2) = e^(-λ) * ( λ^0/0! + λ^1/1! + λ^2/2!)
           = e^(-10) * (1 + 10+ 50/2)
           = 0.0028 (rounded to 4 decimal places)

Thus, the probability of no more than 2 customers arriving in a 5-minute period is 0.0028 or approximately 0.3%. So, the correct choice is (C).

Question 6

A company's manufacturing process uses 500 litres of water at a time. A "scrubbing" machine then removes most of a chemical pollutant before pumping the water into a nearby lake. To meet federal regulations the treated water must not contain more than 80 parts per million (ppm) of the chemical. Because there is a fine charged if regulations are not met, the company sets the machine to attain an average of 75 ppm in the treated water. The machine's output can be described by a Normal model with a standard deviation 4.2 ppm.
a. What percent of the batches of water discharged exceed the 80 ppm standard?
b. The company's lawyers insist that not more than 2% of the treated water should be over the limit. In order to achieve this, to what mean should the company set the scrubbing machine? Assume the standard deviation does not change.

Accepted Answer

The answer of A company's manufacturing process uses 500 litres...

Question 7

The owner of a small convenience store is trying to decide whether to discontinue selling magazines. He suspects that only 5% of the customers buy magazines and thinks that he might be able to sell something more profitable. Before making a final decision, he keeps track of the number of customers who buy magazines on a given day.
a. On one day he had 280 customers. Assuming this day was typical, what would be the mean and standard deviation of the number of customers who buy magazines each day?
b. Surprised by the high number of customers who purchased magazines that day, the owner decided that his 5% estimate must have been too low. How many magazine sales would it have taken to convince you? Justify your answer.

Accepted Answer

The answer of The owner of a small convenience store...

Question 8

A company believes that there is a 25% chance of making a daily profit of $500, a 35% chance that it will be $1000 and 40% chance that it will be $1500.
a. Find the expected value.
b. Find the standard deviation.

Accepted Answer

The answer of A company believes that there is a...

Question 9

Consider the following to answer the question(s) below:
A company's manufacturing process uses 500 litres of water at a time. A "scrubbing" machine then removes most of a chemical pollutant before pumping the water into a nearby lake. To meet federal regulations the treated water must not contain more than 80 parts per million (ppm) of the chemical. Because there is a fine charged if regulations are not met, the company sets the machine to attain an average of 75 ppm in the treated water. The machine's output can be described by a Normal model with a standard deviation 4.2 ppm.
-What percent of the batches of water discharged exceed the 80 ppm standard?
A) 88.3%
B) 11.7%
C) 61.7%
D) 38.3%
E) -11.7%

Accepted Answer

To find the percentage of batches exceeding 80 ppm, we calculate the z-score for 80 ppm given the mean of 75 ppm and the standard deviation of 4.2 ppm. The z-score formula is $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value of interest (80 ppm), $\mu$ is the mean (75 ppm), and $\sigma$ is the standard deviation (4.2 ppm). Plugging in the values, we get $z = \frac{(80 - 75)}{4.2} = \frac{5}{4.2} \approx 1.19$. Looking up 1.19 in a standard normal distribution table or using a calculator, we find that the area to the right of this z-score (which represents the percentage of batches exceeding 80 ppm) is approximately 11.7%.

Question 10

Consider the following to answer the question(s) below:
At a local manufacturing plant, employees must complete new machine set ups within 30 minutes. New machine set-up times can be described by a Normal model with a mean of 22 minutes and a standard deviation of four minutes.
-What percent of new machine set ups take more than 30 minutes?
A) 97.72%
B) 47.72%
C) 2.28%
D) 52.28%
E) 50%

Accepted Answer

We need to find the area to the right of 30 minutes on the Normal distribution curve with the given mean and standard deviation. We can find this using a standard Normal distribution table or calculator, converting the value of 30 minutes to a standardized score:

z = (30 - 22) / 4 = 2

Looking at the standard Normal distribution table, we can find that the area to the right of z = 2 is 0.0228, or 2.28%. Therefore, choice C is the correct answer.

Question 11

Consider the following to answer the question(s) below:
An airline has found that its flights between Toronto and Vancouver can take anywhere from 4.5 - 7.5 hours, with any time between this range being equally likely.
-What is the probability that the flight will take more than 7 hours?
A) 0.83
B) 0.33
C) 0.67
D) 0.17
E) Anywhere between 0% and 100%.

Accepted Answer

The probability of the flight taking more than 7 hours is equal to the probability of the flight taking between 7 and 7.5 hours, which is half of the probability of the entire range between 4.5 and 7.5 hours. Therefore, P(flight takes more than 7 hours) = 0.5*(7.5-7)/(7.5-4.5) = 0.17.

Question 12

Consider the following to answer the question(s) below:
A company believes that there will be a 50% chance of making a profit of $1,000, a 30% chance of making a profit of $1500 and a 20% chance of making a profit of $2,000.
-What is the expected profit?
A) $1,000
B) $300
C) $2,000
D) $1,350
E) $4,500

Accepted Answer

To find the expected profit, we multiply the probability of each profit by its corresponding amount, then add them together. 
Expected Profit = (0.5 x 1000) + (0.3 x 1500) + (0.2 x 2000) = 500 + 450 + 400 = $1,350. Therefore the answer is D.

Question 13

Consider the following to answer the question(s) below:
The number of clients arriving at a bank machine is Poisson distributed with an average of 2 per minute. For a 5 minute period,
-Find the expected value.
A) 2
B) 1.41
C) 3.16
D) 10
E) The expected value for 5 minutes cannot be determined with the given information.

Accepted Answer

The expected value for a Poisson distribution is given by the rate (λ) multiplied by the time period. Here, λ = 2 clients/minute and the time period is 5 minutes, so the expected value is 2 * 5 = 10 clients.

Question 14

A shareholder believes that in one year, there is a 20% chance that his stock will be worth $75, a 50% chance that it will be worth $100, and a 30% chance that it will be worth $140.
a. Find the stock's expected value in one year.
b. Find the standard deviation of the stock's worth in one year.

Accepted Answer

The answer of A shareholder believes that in one year,...

Question 15

The number of plants that are found in a field averages 2 per square metre and follows a Poisson distribution. Based on this,
a. What is the expected value?
b. What is the probability that a 1-square metre section will contain less than 3 plants?.

Accepted Answer

The answer of The number of plants that are found...

Question 16

Consider the following to answer the question(s) below:
An airline has found that its flights between Toronto and Vancouver can take anywhere from 4.5 - 7.5 hours, with any time between this range being equally likely.
-What is the standard deviation?
A) 0.75 hours
B) 0.87 hours
C) 6 hours
D) 0.33 hours
E) The standard deviation cannot be determined.

Accepted Answer

The range of possible flight times is 7.5-4.5=3 hours. The midpoint is (7.5+4.5)/2=6 hours. We can use the formula for standard deviation, sigma, which is given by: sigma=sqrt((b-a)^2/12), where a is the lower limit, b is the upper limit, and sqrt means "square root". Plugging in the values we get: sigma=sqrt((7.5-4.5)^2/12)=sqrt(3^2/4)=sqrt(9/4)=0.75. Therefore, the standard deviation is 0.75 hours, which corresponds to choice B.

Question 17

At a local manufacturing plant, employees must complete new machine set ups within 30 minutes. New machine set-up times can be described by a Normal model with a mean of 22 minutes and a standard deviation of four minutes.
a. What percent of new machine set ups take more than 30 minutes?
b. The typical worker needs five minutes to adjust to their surroundings before beginning their duties. What percent of new machine set ups are completed within 25 minutes to allow for this?

Accepted Answer

The answer of At a local manufacturing plant, employees must...

Question 18

Consider the following to answer the question(s) below:
At a local manufacturing plant, employees must complete new machine set ups within 30 minutes. New machine set-up times can be described by a Normal model with a mean of 22 minutes and a standard deviation of four minutes.
-The typical worker needs five minutes to adjust to their surroundings before beginning their duties. What percent of new machine set ups are completed within 25 minutes to allow for this?
A) 77.3%
B) 27.3%
C) 22.7%
D) 72.7%
E) 0.75%

Accepted Answer

To find the percent of new machine set-ups completed within 25 minutes, we need to use the Z-score formula:

Z = (X - μ) / σ

where X is the time (25 minutes), μ is the mean (22 minutes), and σ is the standard deviation (4 minutes).

Z = (25 - 22) / 4 = 0.75

Looking up the Z-score in a standard normal distribution table, we find that the probability of a Z-score less than or equal to 0.75 is 0.7734.

Therefore, the percentage of new machine set-ups completed within 25 minutes is approximately 77.3% (0.7734 x 100%).

Question 19

The number of calls received by a call centre follows a Poisson distribution with an average of 5 calls per minute. For a 1 minute period,
a. Find the expected value.
b. Find the probability of receiving no more than 2 calls.

Accepted Answer

The answer of The number of calls received by a...

Question 20

Consider the following to answer the question(s) below:
A courier service claims that no more than 5 % of all of its deliveries arrive late. Assuming deliveries are independent, a sample of 10 deliveries is randomly selected.
-What is the probability that exactly 2 of the sample deliveries arrives late?
A) 0.0746
B) 0.9884
C) 0.2347
D) 0.0439
E) 0.1125

Accepted Answer

The probability of exactly 2 deliveries arriving late can be calculated using the binomial probability formula: P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where n=10, k=2, and p=0.05. This results in approximately 0.0746.

Quiz 9: Random Variables and Probability Distributions

Consider the following to answer the question(s) below: A company believes that there will be a 50% chance of making a profit of $1,000, a 30% chance of making a profit of $1500 and a 20% chance of making a profit of $2,000. -Find the standard deviation for the company's profit?

Consider the following to answer the question(s) below: The number of clients arriving at a bank machine is Poisson distributed with an average of 2 per minute. For a 5 minute period, -What is the probability that no more than 2 customers will arrive in a 5 minute period?

A company believes that there is a 25% chance of making a daily profit of $500, a 35% chance that it will be $1000 and 40% chance that it will be $1500. a. Find the expected value. b. Find the standard deviation.

Consider the following to answer the question(s) below: An airline has found that its flights between Toronto and Vancouver can take anywhere from 4.5 - 7.5 hours, with any time between this range being equally likely. -What is the probability that the flight will take more than 7 hours?

Consider the following to answer the question(s) below: A company believes that there will be a 50% chance of making a profit of $1,000, a 30% chance of making a profit of $1500 and a 20% chance of making a profit of $2,000. -What is the expected profit?

Consider the following to answer the question(s) below: The number of clients arriving at a bank machine is Poisson distributed with an average of 2 per minute. For a 5 minute period, -Find the expected value.

A shareholder believes that in one year, there is a 20% chance that his stock will be worth $75, a 50% chance that it will be worth $100, and a 30% chance that it will be worth $140. a. Find the stock's expected value in one year. b. Find the standard deviation of the stock's worth in one year.

The number of plants that are found in a field averages 2 per square metre and follows a Poisson distribution. Based on this, a. What is the expected value? b. What is the probability that a 1-square metre section will contain less than 3 plants?.

Consider the following to answer the question(s) below: An airline has found that its flights between Toronto and Vancouver can take anywhere from 4.5 - 7.5 hours, with any time between this range being equally likely. -What is the standard deviation?

The number of calls received by a call centre follows a Poisson distribution with an average of 5 calls per minute. For a 1 minute period, a. Find the expected value. b. Find the probability of receiving no more than 2 calls.