Question 1

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

a. E(X) = Σ xP(x) = $75(0.20) + $100(0.50) + $140(0.30) = $107
b. σ = SD(X) =  , Var(X) = Σ (x - μ)2 P(x)
Var(X) = Σ(x - μ)2 P(x) = (75 - 107)2(0.20) + (100 - 107)2(0.50) + (140 - 107)2(0.30) = 556
σ = SD(X) =  =  = $23.58

Question 2

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

a. The expected value of the Poisson distribution is E(X) = λ = 2.
b. P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = e-2  = 0.6767

Question 3

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

a. P(x > 30) = P  = P(z > 2.00) = 0.0228 = 2.28%
b. P(x ≤ 25) = P  = P(z ≤ 0.75) = 0.7734

Question 4

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) 78.63 ppm
C) 71.39 ppm
D) 80 ppm
E) 60 ppm

Accepted Answer

We need to find the mean of the Normal distribution that the machine produces, so that not more than 2% of the water is over the limit of 80 ppm. We can use the z-score formula:

z = (x - mean) / standard deviation

where x is the limit of 80 ppm, and we want to find the mean that corresponds to a z-score of 2.05 (since we want the top 2% of the distribution to be over the limit):

2.05 = (80 - mean) / 4.2

Solving for mean, we get:

mean = 71.39

Therefore, the company should set the scrubbing machine to attain an average of 71.39 ppm in the treated water in order to meet the lawyers' requirement that not more than 2% of the water should be over the limit of 80 ppm.

Question 5

The number of calls received by a call centre follows the 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 6

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 to 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) 0.5 hours

Accepted Answer

Since the range of possible flight times is 7.5 - 4.5 = 3 hours, and any time within this range is equally likely, we can use the formula for the standard deviation of a uniform distribution:

SD = (b-a)/sqrt(12)

where a is the lower bound of the range (4.5 hours) and b is the upper bound (7.5 hours).

SD = (7.5 - 4.5)/sqrt(12) = 0.866 hours, which is closest to option B, 0.87 hours.

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

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 to 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) 0.99

Accepted Answer

The probability of the flight taking more than 7 hours is equal to the probability of the flight taking any time between 7.0 and 7.5, which is 0.5/(7.5-4.5) = 0.17, as all times within the range are equally likely. Therefore, the answer is D, 0.17.

Question 9

Consider the following to answer the question(s) below:
A courier service claims that only 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 arrive late?
A) 0.0746
B) 0.9884
C) 0.2347
D) 0.0439
E) 0.1125

Accepted Answer

This is a binomial probability problem where n = 10 (number of trials), k = 2 (number of successes), and p = 0.05 (probability of success on a single trial). The formula for binomial probability is P(X = k) = (nCk) * p^k * (1-p)^(n-k). Plugging in the values: P(X = 2) = (10C2) * (0.05)^2 * (0.95)^8 = 0.0746.

Question 10

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) $1,350

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 11

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 12

Consider the following to answer the question(s) below:
A courier service claims that only 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) 0.0115
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 = 0.2993. The probability of 2 deliveries arriving late is 45 * 0.05^2 * 0.95^8 = 0.1020. The probability of more than 2 deliveries arriving late is 1 - 0.5987 - 0.2993 - 0.1020 = 0.0115.

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,
-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) 0.1353

Accepted Answer

The answer of Consider the following to answer the question(s)...

Question 14

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) 2.24

Accepted Answer

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

Question 15

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

The answer of The owner of a pet store is...

Question 16

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 17

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

We can calculate the expected profit by multiplying the probability of each profit amount by its corresponding profit and then summing the results:
Expected profit = (0.5 * 1000) + (0.3 * 1500) + (0.2 * 2000) = 500 + 450 + 400 = $1,350. Therefore, the best choice is D.

Question 18

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 19

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 probability that a new machine set-up takes more than 30 minutes. We can standardize the value of 30 using the formula z = (x - μ)/σ, where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.

z = (30 - 22)/4 = 2

Looking at the z-table, we can see that the area to the right of 2 is 0.0228, which is equal to 2.28% (rounded to two decimal places). This means that only 2.28% of new machine set-ups take more than 30 minutes, so the answer is choice C, 2.28%.

Question 20

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%.

Quiz 9: Random Variables and Probability Distributions

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?

The number of calls received by a call centre follows the 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.

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 to 7.5 hours, with any time between this range being equally likely. -What is 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 to 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 courier service claims that only 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 arrive late?

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: A courier service claims that only 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?

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?

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.

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 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.