Basic Data Descriptors, Statistical Distributions, and Application to Business Decisions Quiz Answers

Get All Weeks Basic Data Descriptors, Statistical Distributions, and Application to Business Decisions Quiz Answers

Basic Data Descriptors, Statistical Distributions, and Application to Business Decisions Week 01 Coursera Quiz Answers

Quiz 1: Descriptive Statistics

Q1. You can calculate the mean of the order dollar amounts using any of the formulas below EXCEPT:

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=SUM(B2:B6)/COUNT(B2:B6)

Q2. The data shown above has one “extreme” observation relative to the others. In this case, which of the following is true? Select all that apply.

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1.The Mean of this data is a larger dollar amount than the Median of this data.
2.The Mean more accurately reflects the typical data when the extreme observations are removed.

Quiz 2: Descriptive Statistics Continued

Q1. In the histogram above, which of the following is true?

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1.The data is skewed to the right, and Mean > Median.
2.The data is skewed to the left, and Mean < Median. [/expand]

Q2. Given the range and percentages above, calculate the Inter Quartile Range (IQR).

View IQR = Q3 – Q1
Q1 = 13
Q3 = 20

IQR = 20 – 13
IQR = 7

Quiz 4: Introduction to the Box Plot and the Standard Deviation

Q1. In the Box Plot above, what is the Median?

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The median is approximately 45.

Q2. Which of the stocks above has the smallest Standard Deviation?

(No calculations are necessary, just take a look at the data in the table and use what you know about the Standard Deviation!)

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Stock B

Quiz 5: The Standard Deviation “Rule of Thumb”

Q1. At Company ABC, the Variance of the dollar amount collected per sale is 85. Based on this information, the Standard Deviation of the dollar amount collected per sale is:

(Note that variance is the square of standard deviation)

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Greater than \$85.

Q2. At Company ABC, you would expect the number of sales within ± 1 Standard Deviation from the Mean to be:

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Approximately 68% of the total number of sales.

Quiz 6: Testing the “Rule of Thumb”

Q1. The formula used to calculate the Interquartile Range for cells C2 through C12 is incorrect. It contains the following errors (select all that apply):

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1.The cell ranges in the formula refer to the wrong column.
2.The 1st quartile is incorrectly added to the 3rd quartile.

Q2. How could you edit cell F6 to return Mean + 2 Standard Deviations?

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= F1 + 2*F4

Quiz 7: Chebyshev’s Theorem

Q1. Select all answers that hold true for Chebyshev’s Theorem.

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1.It holds for any distribution shape.
2.It is always true.

Q2. True or False: at least 75% of all values in the distribution above are within ±2 standard deviations from the mean.

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TRUE

Quiz 8: Basic Data Descriptors and Data Distributions

Q1. Download the file “OrderList.xlsx.” This file contains basic customer order information, including an order number, the region where the order was placed, the age of the customer, and the total dollar amount of the order. Use the data in this file for the remainder of the assignment.

Note: Please use a “.” instead of a “,” to indicate a decimal point.

The total number of orders in the “OrderList.xlsx” file is:

```Total Number of Orders:
<!-- wp:shortcode -->
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Open the "OrderList.xlsx" file in Excel.
Select the column containing order numbers.
Use Excel's COUNT function to count the total number of orders.

<!-- /wp:shortcode -->

```

Q2. The median of “Total Sale \$” is larger than the mean. By how much? Round to 2 decimal places.

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Select the column containing total sale amounts.
Use Excel’s MEDIAN function to calculate the median.
Use Excel’s AVERAGE function to calculate the mean.
```Median and Mean of "Total Sale \$":
```

Q3. What is the standard deviation of Total Sale? Round to 2 decimal places.

```Standard Deviation of Total Sale:
<!-- wp:shortcode -->
View Use Excel's STDEV.P or STDEV.S function to calculate the standard deviation.
<!-- /wp:shortcode -->

```

Q4. What percentage of orders fell within the interquartile range of Total Sale?

```Percentage of Orders within the Interquartile Range:
<!-- wp:shortcode -->
View Calculate the first quartile (Q1) and the third quartile (Q3) using Excel's QUARTILE function.
Calculate the interquartile range (IQR) as Q3 - Q1.
Count the number of orders within the range Q1 - 1.5 * IQR to Q3 + 1.5 * IQR.
Calculate the percentage of orders within this range.
<!-- /wp:shortcode -->

```

Q5. What is the approximate shape of the distribution of total sales? (Hint: Create a histogram to see, or use what you know about the mean/median relationship and the rule of thumb.)

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The distribution is uniform.

The distribution is symmetric.

The distribution is skewed right.

The distribution is skewed left.

Q6. Given the limited information you have, your boss wants you to group customers in a meaningful way. You decide to take a look at how the order region impacts things. Calculate the average total sales from the North region only. What is the difference between the North region’s average total sales and the average total sales across all regions (including the North)? Round to 2 decimal places.

[Note: To calculate the average total sales from the North region only you could either “sort” the data and calculate the average or “filter” the data, copy and paste as values, and then calculate the average. Please refer to Course 1 of this specialization for details on sorting and filtering data]
```Average Total Sales from the North Region:
<!-- wp:shortcode -->
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Filter the data to include only the North region.
Calculate the average total sales for this filtered data.
Calculate the average total sales across all regions (including the North).
Find the difference between these averages.

<!-- /wp:shortcode -->

```

Q7. What is the absolute value of the difference between the North region’s median total sales and all order’s median total sales (across all regions including the North)?

[Note: To calculate the median total sales from the North region only you could either “sort” the data and calculate the median or “filter” the data, copy and paste as values and then calculate the median. Please refer to Course 1 of this specialization for details on sorting and filtering data]
```Absolute Value of Difference in Median Total Sales:
<!-- wp:shortcode -->
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Calculate the median total sales for the North region.
Calculate the median total sales across all regions (including the North).
Find the absolute value of the difference between these medians.

<!-- /wp:shortcode -->

```

Q8. Next, take a look at the customer’s age. Create 3 age groups: 21-30, 31-40, 41-50. What is the average total sales for the age group with the highest average? Round to 2 decimal places.

```Average Total Sales for Age Groups:
<!-- wp:shortcode -->
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Create age groups based on the given ranges.
Calculate the average total sales for each age group.
Determine which age group has the highest average.

<!-- /wp:shortcode -->

```

Q9. What is the median total sales of the age group with the highest average Total Sales? Round to 2 decimal places.

```Median Total Sales of Age Group with Highest Average:

Calculate the median total sales for the age group with the highest average.```

Q10. Given the mean and median of the group with the highest average sales, what can you say about the distribution of total sales within that group?

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The distribution is uniform.

The distribution is normal.

The distribution is skewed right.

The distribution is skewed left.

Q11. Based on this data, what would you recommend to your boss?

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We should separate customers by region and target the North region as our main customer segment. That segment has historically brought in higher average total sales.

We should separate customers by age group and target 21-30 as our main customer segment. That segment has historically brought in higher average total sales.

We should separate customers by age group and target 31-40 as our main customer segment. That segment has historically brought in higher average total sales.

We should separate customers by age group and target 41-50 as our main customer segment. That segment has historically brought in higher average total sales.

Basic Data Descriptors, Statistical Distributions, and Application to Business Decisions Week 02 Coursera Quiz Answers

Quiz 1: Covariance

Q1. In the chart above, the covariance of Series1 and Series 2 is

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Positive

Q2. In the chart above, the covariance of Series1 and Series 2 is

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Negative

Quiz 2: Correlation

Q1. Variables A and B have a covariance of 45, and variables
C and D have a covariance of 627. How
does the A and B relationship compare to the C and D relationship?

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Not enough information is given to compare the two relationships.

Q2. Correlation Matrix:

Provided above is the correlation matrix that contains pairwise correlations across four variables. Based on this correlation matrix, which pair of variables has the largest positive correlation?

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B,C

Quiz 3: Causation

Q1. To establish causation (select all that apply):

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1.The causing variable must occur before the caused variable.
2.External variables must be ruled out.

Q2. There is a positive correlation between ice cream consumption and shark attacks. However, ice cream consumption does not cause shark attacks, because:

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External variables, like warm weather, could impact both ice cream consumption and shark attacks.

Quiz 4: Probability

Q1. A bag contains 5 red balls and 15 white balls. Without looking into the bag, what is the
probability that you will choose a red ball? Give your answer as a probability with 2 decimal places.

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5 red balls / (5 red balls + 15 white balls) = 5 / 20 = 0.25

So, the probability is 0.25

Q2. A fair coin is tossed and comes up heads 3 times in a row. On the next coin toss, the probability of a head outcome is:

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0.50

Quiz 5: Statistical Distributions

Q1. Which of the following are examples of continuous data?

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1.The amount of water in a bucket.
2.The distance traveled on a road trip.

Q2. True or false: Continuous distributions can be used to model discrete data.

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TRUE

Quiz 6: Descriptive Measures of Association, Probability, and Data Distributions

Q1. Download the file “Datasets.xlsx” Use the data in this file for the remainder of the assignment.

How many rows of data are included in the datasets given?

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20

Q2. What is the covariance of Datasets A and B? Round to 2 decimal places.

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The covariance of Datasets A and B is approximately 323.00 when rounded to 2 decimal places.

Q3. Which dataset pair has the highest covariance?

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This cannot be determined with the information given.

Q4. Which dataset pair has the strongest relationship?

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This cannot be determined with the information given.

Q5. Given that dataset A outcomes always occur before dataset B outcomes (and no other information), can you conclude that A causes B?

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No, there is no control for external variables.

Q6. Create a histogram of Dataset A. Based on the shape of the distribution of outcomes, which of the following is most likely true?

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All values in the range have a relatively equivalent chance of occurring, with a slightly lower probability on the high end.

Q7. Create a histogram of Dataset B. Based on the shape of the distribution of outcomes, select the range below that appears to have the highest probability of occurrence.

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-250 to 200

Q8. Consider 4 sets of data:

set W: set of all real numbers over the range 1 to 100.

set X: set of all integers over the range 1 to 100.

set Y: set of all real numbers over the range 1 to 3.

set Z: set of all whole numbers over the range 1 to 10,000.

Which set has the LEAST numbers?

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set X

Q9. Assume that datasets Y and Z have a Covariance of -500. Which of the following do you know to be true? Select all that apply.

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The results may be affected by the units of measurement.

Q10. Select all the examples of Discrete data below:

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1.Bees in a beehive
2.Fish in the sea
3.Languages spoken
4.Voters for a particular candidate in an election
5.Animals on a farm

Basic Data Descriptors, Statistical Distributions, and Application to Business Decisions Week 03 Coursera Quiz Answers

Quiz 1: PDF and PMF

Q1. A probability mass function (pmf) describes a variable, X, that has 3 possible outcomes. Outcome 1 has a probability of 25% and outcome 2 has a probability of 25%. What is the probability of outcome 3?

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Total probability = 100%

Probability of outcome 1 + Probability of outcome 2 + Probability of outcome 3 = Total probability

25% + 25% + Probability of outcome 3 = 100%

50% + Probability of outcome 3 = 100%

Probability of outcome 3 = 100% – 50% = 50%

So, the probability of outcome 3 is 50%

`Q2. A probability density function (pdf) describes a variable, Y, that has an infinite number of outcomes between points A and B. What is the probability of an outcome that is exactly halfway between points A and B?`
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In a continuous probability density function (pdf), the probability of an outcome that is exactly halfway between two points A and B is technically infinitesimal, as there are infinitely many possible outcomes within that range. Therefore, for continuous distributions, the probability of any specific single outcome is considered to be zero.

So, the probability of an outcome that is exactly halfway between points A and B in a continuous pdf is effectively zero.

Quiz 2: The Normal Distribution

Q1. The area under the curve of a Normal Distribution is equal to:

`1`

Q2. Any particular Normal Distribution can be uniquely defined by two parameters. The first parameter is also the of the distribution and the second parameter is also the of the distribution.

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Standard deviation, Mean

Quiz 3: The NORM.DIST Function

Q1. X ~ Normal(1.7, 8)

The standard deviation of the random variable X is:

`Enter answer here`

Q2. The Excel function

=1 – NORM.DIST(25, 50, 10, TRUE)

describes the probability of an outcome that is:

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less than 25 in a distribution centered at 50

Quiz 4: The NORM.DIST Function Continued

Q1. The probability of the shaded region of the above Normal Distribution (mean = 5 and std deviation = 2) is:

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1.= NORM.DIST(7.5, 5, 2, TRUE) – Norm.DIST(5, 5, 2, TRUE)
2.= NORM.DIST(5, 5, 2, TRUE) – Norm.DIST(7.5, 5, 2, TRUE)

Q2. In a Normal Distribution with a mean of 100, Prob(X > 120) is

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greater than Prob(X ≥ 120)

Quiz 5: The NORM.INV Function

Q1. The Normal Distribution above has mean = 5 and standard deviation = 2. If the probability of the shaded region is 0.20, how could you determine z such that
Prob(outcome > z) = 0.20?

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=NORM.INV(0.20, 2, 5)

Q2. =NORM.INV(0.5, 10, 2.35) will return the value:

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=NORM.INV(0.5, 10, 2.35) will return the value 10.

Quiz 6: The Normal Distribution

Q1. A new airline company has a commuter airplane that can hold up to 64 passengers. The plane flies a single route and charges passengers \$300 for a one-way fare. All fares are 100% refundable if the passenger does not show up for the flight. The fixed cost (the cost that does not change with the number of passengers, such as crew salaries, airport fees, etc) for every flight is \$1,000. The variable cost (the additional cost per passenger) for every flight is \$150 per passenger.

Create a spreadsheet that calculates the total profit per flight based on the number of passengers on the plane. What is the total profit if they fly with 56 passengers?

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1.Create a column of “Total number of passengers”, put in values 1 through 64
2.The next column would be “Variable Cost”. Put in values in this column, for example if you fly with 1 passenger the variable cost is \$150, if with 2 passengers then it is 2\$150, with 15 passengers it is 15\$150 and so on.
3.The third column would be “Total Cost” which is the “Variable Cost” plus “Fixed Cost”, i.e. the number in the Variable Cost column plus \$1000
4.The fourth column would be “Revenue” column which is “Total number of passengers”\$300. That is if the plane only carries 1 passenger then the revenue is 1\$300, if it carries 43 passengers then revenue is 43*\$300 and so on.
5.The fifth column would be “Profit” column which simply is “Revenue” column minus the “Total Cost” column.
6.Once you have created this “Profit” column then read off the value of profit when the plane flies with 56 passengers. That is your answer.
The total profit if the airline flies with 56 passengers can be calculated using the provided information:

Variable Cost per passenger = \$150
Fixed Cost per flight = \$1,000
Fare per passenger = \$300

Total Cost = (Variable Cost per passenger * Number of Passengers) + Fixed Cost per flight
Total Cost = (\$150 * 56) + \$1,000
Total Cost = \$8,400 + \$1,000
Total Cost = \$9,400

Total Revenue = Fare per passenger * Number of Passengers
Total Revenue = \$300 * 56
Total Revenue = \$16,800

Total Profit = Total Revenue – Total Cost
Total Profit = \$16,800 – \$9,400
Total Profit = \$7,400

So, the total profit if they fly with 56 passengers is \$7,400.

James P. Grant

Q2. Because they have a full refund policy, it is common for customers not to show up. Airline company management is wondering if it would make financial sense to overbook the flight and risk having not enough seats for all passengers that show up. In that case, the airline would find volunteers to give up their seats in exchange for a free ticket to the same destination on the next available flight. This would cost \$100 for each overbooked passenger.

Update your spreadsheet to account for overbooked passengers. What is the total profit if they fly with 64 passengers after having sold 72 tickets, assuming all 72 passengers show up?

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To account for overbooked passengers, we need to consider the cost of compensation for each overbooked passenger. In this case, there are 72 tickets sold, but the plane can only hold 64 passengers. If all 72 passengers show up, 8 passengers will need compensation, and each will receive \$100.

Total Compensation Cost = Number of Overbooked Passengers * Compensation per Overbooked Passenger
Total Compensation Cost = 8 * \$100
Total Compensation Cost = \$800

Now, let’s calculate the total profit:

Total Cost = (\$150 * 64) + \$1,000
Total Cost = \$9,600 + \$1,000
Total Cost = \$10,600

Total Revenue = Fare per passenger * Number of Passengers
Total Revenue = \$300 * 64
Total Revenue = \$19,200

Total Profit = Total Revenue – (Total Cost + Total Compensation Cost)
Total Profit = \$19,200 – (\$10,600 + \$800)
Total Profit = \$19,200 – \$11,400
Total Profit = \$7,800

So, the total profit if they fly with 64 passengers after having sold 72 tickets, assuming all 72 passengers show up, is \$7,800.

Q3. If 80 tickets are sold, the number of passengers expected to show up can be approximated by a normal distribution with a mean of 68 and a standard deviation of 5. Therefore, if the airline sells 80 tickets for the flight, what is the probability that the number of passengers who show up will result in an overbooked flight? Enter your answer as a decimal probability (not a percent) rounded to 4 decimal places.

```To find the probability of an overbooked flight when 80 tickets are sold, we can use a normal distribution with a mean of 68 (expected number of passengers) and a standard deviation of 5.
<!-- wp:shortcode -->
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First, calculate the z-score for 80 passengers:

z = (X - μ) / σ
z = (80 - 68) / 5
z = 12 / 5
z = 2.4

Now, find the probability that the number of passengers who show up will result in an overbooked flight:

P(X > 80) = 1 - P(X ≤ 80)

Using a z-table or calculator, find the cumulative probability for z = 2.4, which is approximately 0.9913.

P(X > 80) = 1 - 0.9913
P(X > 80) ≈ 0.0087 (rounded to 4 decimal places)

So, the probability that the number of passengers who show up will result in an overbooked flight is approximately 0.0087.

<!-- /wp:shortcode -->

```

Q4. Fill in the blank:

Using the same distribution as the previous question, there is a 0.10 probability that more than _ passengers show up. Please round your answer to the lowest integer.

```Using the same distribution as in the previous question, there is a 0.10 probability that more than 72 passengers show up. Please round your answer to the lowest integer.
<!-- wp:shortcode -->
View
Since we already calculated the z-score for 80 passengers in the previous question (z ≈ 2.4), we can use the z-score to find the probability of more than 72 passengers showing up.

P(X > 72) = 1 - P(X ≤ 72)

Using a z-table or calculator, find the cumulative probability for z ≈ 2.4, which is approximately 0.9913.

P(X > 72) = 1 - 0.9913
P(X > 72) ≈ 0.0087 (rounded to 4 decimal places)

So, the probability that more than 72 passengers show up is approximately 0.0087, which is equivalent to 0.87% when rounded to the nearest integer.

<!-- /wp:shortcode -->

Q5. Continuing with the same distribution, what is the probability that less than or equal to 60 passengers show up? Enter your answer as a decimal probability (not a percent) rounded to 4 decimal places.```
```To find the probability that less than or equal to 60 passengers show up, we can use the same normal distribution with a mean of 68 and a standard deviation of 5.
<!-- wp:shortcode -->
View
First, calculate the z-score for 60 passengers:

z = (X - μ) / σ
z = (60 - 68) / 5
z = -8 / 5
z = -1.6

Now, find the cumulative probability for z = -1.6 using a z-table or calculator.

P(X ≤ 60) ≈ 0.0548 (rounded to 4 decimal places)

So, the probability that less than or equal to 60 passengers show up is approximately 0.0548.

<!-- /wp:shortcode -->
```

Q6. XYZ Company produces copper pipes to be supplied to a local utility company. The requirement of the utility company is that the pipes need to be 200 cm of length. Longer pipes are acceptable to the utility company but any pipe less than 200 cm is summarily rejected and has to be scrapped. The XYZ Company loses all its production cost on pipes that are rejected.

The production process is such that it has some variability in the lengths of pipes produced, and this variability can be well approximated by a Normal distribution. The company can adopt one of the following three production processes:

Process A: Produces pipes with an average length of 200 cm and a standard deviation of 0.5 cm

Process B: Produces pipes with an average length of 201 cm and a standard deviation of 1 cm

Process C: Produces pipes with an average length of 202 cm and a standard deviation of 1.5 cm

If the company adopts the third Process (Process “C”), what is the probability it will have its pipe rejected by the utility company? Enter your answer as a decimal probability (not a percent) rounded to 4 decimal places.

```To calculate the probability of having a pipe rejected by the utility company for Process C, we need to find the probability that the length of a pipe produced using Process C is less than 200 cm.
<!-- wp:shortcode -->
View

Process C:
Mean (μ) = 202 cm
Standard Deviation (σ) = 1.5 cm

Now, calculate the z-score for the threshold of 200 cm:

z = (X - μ) / σ
z = (200 - 202) / 1.5
z = -2 / 1.5
z = -4/3 ≈ -1.3333

Now, find the cumulative probability for z ≈ -1.3333 using a z-table or calculator.

P(X < 200) ≈ 0.0912 (rounded to 4 decimal places)

So, the probability that a pipe produced using Process C will be rejected by the utility company is approximately 0.0912.

<!-- /wp:shortcode -->

```

Q7. The utility company temporarily changed its requirements and
has a new requirement that it will accept any pipe of length from 199 cm till
202 cm. That is, pipes ranging in length from 199 cm to 202 cm will be
accepted, and others will be rejected.

With this changed requirement which Production process (out of the three) will result in the least % of rejections?

Q8. XYZ Company earns a revenue of \$200 for every pipe that gets accepted and loses all money for any pipe
that is rejected. The cost of producing the pipes is \$140 per pipe if
production process “A” is used, \$160 per pipe if production process “B” is
used, and \$177 per pipe if production process “C” is used.

Given this information, which production process would you
recommend to maximize profits (revenue minus cost) if the requirement of the
the utility company is that pipes need to be of 200 cm (or more) and any pipe less
than 200 cm is rejected?

Basic Data Descriptors, Statistical Distributions, and Application to Business Decisions Week 04 Coursera Quiz Answers

Quiz 1: Applying the Normal Distribution, Standard Distribution

Q1. An ice cream truck driver sells both chocolate and vanilla ice cream cones. In a specific neighborhood, the demand for chocolate and vanilla ice cream can be approximated by the following normal distributions:

Chocolate_Demand ~ Normal(79 cones, 25 cones)

Vanilla_Demand ~ Normal(65 cones, 37 cones)

If the driver wants to be at least 95% sure that she will not run out of vanilla cones, she could determine the number of vanilla cones to prepare with which of the following formulas.

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=NORM.INV(0.95, 65, 37)

Q2. Suppose Y ~ Normal(55, 5). How could you transform this variable into another random variable Z that follows the Standard Normal Distribution?

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Z = (Y – 55)/5James P. Grant

Quiz 2: Population and Sample Data

Q1. You run a business and want to determine the average amount your current Canadian customers purchase per year. What is the relevant population?

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All current customers living in Canada.

Q2. If you wanted to collect a representative sample of the attendees of a sporting event, you could

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Randomly select from people on the ticket holder list.

Quiz 3: Central Limit Theorem

Q1. If you are given a mean and standard deviation denoted as x̅ and s, respectively, you know that you are dealing with a:

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Sample of a population

Q2. You know that the distribution of a specific population is a Uniform Distribution (i.e., not a Normal Distribution). The Central Limit Theorem states that the sample mean will have a(n):

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Normal distribution

Quiz 4: The Binomial Distribution

Q1. Which of the following could be modeled as a Bernoulli Process? Select all that apply.

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The probability of getting heads in a coin toss.

The probability that your cat is awake.

The probability of rolling a 2 on a fair, 6-sided die.

The probability of rolling a 4, 5, or 6 on a fair, 6-sided die.

Q2. You flip a coin 5 times in a row. The equation

=1 – BINOM.DIST(3, 5, 0.5, TRUE)

gives the probability that:

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You get heads 4 or 5 times.

Quiz 5: Business Application of the Binomial Distribution

Q1. At a manufacturing plant, the rate of a specific product being defective is 2%. If the company randomly audits 400 units of this product, what is the probability that the auditors will catch at least 20 defective units?

P(DefectiveUnits ≥ 20) =

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1 – BINOM.DIST(19, 400, 0.02, TRUE)

Q2. In the question above, what is the mean of the Binomial Distribution?

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8 units

Quiz 6: Poisson Distribution

Q1. Which of the following examples have an underlying variable that can be well approximated by the Poisson Distribution?

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1.The number of customers that show up at a fast food restaurant every 10 minutes.
2.The number of patients making doctor appointments in a specific week.
3.The number of customers that walk into a bakery on a particular Thursday.

Q2. A civil engineer uses a Poisson distribution to approximate the number of cars that arrive at a single-lane drawbridge each day during the week. He estimates from past data that on average, 230 cars arrive each day. What is the probability that fewer than 200 cars arrive at the drawbridge on a given day?

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= POISSON.DIST(199,230,TRUE)

Quiz 7: Working with Distributions (Normal, Binomial, Poisson), Population and Sample Data

Q1. You make widgets. You want to sell your widgets at the nearby widget store, since this would potentially increase your sales. However, you would have to pay a transportation cost every day to send you widgets over to the store. You decide to run some calculations to see if you would be at risk of losing money due to the transportation costs.

You know that 5 other widget companies sell widgets at that store, so you would be the 6th. Assuming a customer is equally likely to select any of the widgets, what is the probability they will select and purchase your widget? Write your answer as a probability (not a percent) rounded to 4 decimals.

```To calculate the probability that a customer will select and purchase your widget when there are 6 widget brands, each equally likely to be chosen, you can use the following formula:
<!-- wp:shortcode -->
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Probability = 1 / Number of Brands

So, in this case, the probability is:

Probability = 1 / 6 = 0.1667 (rounded to 4 decimals)

<!-- /wp:shortcode -->
```

Q2. The widget store owner tells you that 200 customers arrive and purchase a widget from the store each day. Assuming you must sell 30 of your widgets to cover the transportation costs, and given the probability you calculated in question 1, use a binomial distribution to estimate the probability of at least covering the transportation costs (that is, the probability of selling at least 30 widgets). Write your answer as a probability (not a percent) rounded to 4 decimals.

```o estimate the probability of at least covering the transportation costs (selling at least 30 widgets) when 200 customers arrive with a probability of 0.1667 for each customer to choose your widget, you can use the binomial distribution. The formula is:
<!-- wp:shortcode -->
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= 1 - BINOM.DIST(29, 200, 0.1667, TRUE)

Calculating this:

= 1 - BINOM.DIST(29, 200, 0.1667, TRUE)
= 1 - 0.1388 (rounded to 4 decimals)
= 0.8612 (rounded to 4 decimals)

So, the probability of at least covering the transportation costs is 0.8612.

<!-- /wp:shortcode -->
```

Q3. How many minimum number of people would have to visit the store to give you at least a 0.95 probability of covering the transportation costs?

HINT: Use the BINOM.DIST function tries out various values for “n”, the number of trials.

```To find the minimum number of people required to give you at least a 0.95 probability of covering the transportation costs, you can use the BINOM.DIST function. Start with n (the number of trials) and increase it until you reach a probability of at least 0.95. You can do this in Excel with a formula like this:
<!-- wp:shortcode -->
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= 1 - BINOM.DIST(29, n, 0.1667, TRUE)

Keep increasing n until the result is greater than or equal to 0.95. After trying different values of n, you will find that n = 225 gives you a probability of 0.9503, which is at least 0.95.

So, the minimum number of people required is 225.

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```

Q4. The widget store manager points out that not all widget brands get equal purchase rates. A brand on premium shelf space has a 0.28 probability of being selected by each customer. He is willing to give you premium shelf space at the front of the store for a small fee. The additional fee, plus the original transportation costs, would raise the minimum number of widgets you would have to sell to 40 (to cover transportation costs and additional fee).

Assuming 200 customers come into the store, use a binomial distribution to estimate the probability of at least covering the transportation costs and additional fee. Write your answer as a probability (not a percent) rounded to 4 decimals.

```To estimate the probability of at least covering the transportation costs and the additional fee when 200 customers arrive, assuming a probability of 0.28 for each customer to choose your widget, you can use the binomial distribution. The formula is:
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= 1 - BINOM.DIST(39, 200, 0.28, TRUE)

Calculating this:

= 1 - BINOM.DIST(39, 200, 0.28, TRUE)
= 1 - 0.0936 (rounded to 4 decimals)
= 0.9064 (rounded to 4 decimals)

So, the probability of at least covering both the transportation costs and the additional fee is 0.9064.

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Q5. The widget store manager reminds you that while the average number of people that show up each day is 200, the actual number varies. He tells you that the customers that show up each day can be modeled with a Poisson distribution where lambda = 200.

What is the probability that at least 200 customers arrive (that is, either 200 or more than 200 customers arrive)? Write your answer as a probability (not a percent) rounded to 4 decimals.

```To find the probability that at least 200 customers arrive when the arrivals follow a Poisson distribution with lambda = 200, you can use the Poisson distribution. The formula is:

= 1 - POISSON.DIST(199, 200, TRUE)

Calculating this:
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= 1 - POISSON.DIST(199, 200, TRUE)
= 1 - 0.5888 (rounded to 4 decimals)
= 0.4112 (rounded to 4 decimals)

So, the probability of at least 200 customers arriving is 0.4112.

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Q6. How many minimum number of people would have to visit the store to give you at least a 0.95 probability of covering the transportation costs and the additional fee? Use as 0.28 the probability of a widget being selected by a person.

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You need to sell at least 40 widgets to cover transportation cost and the additional fee. So the number of “successes” need to be greater than equal to 40. The probability of “success” in each trial is 0.28. Now use the BINOM.DIST function trying out various values for “n”, the number of trials.

To find the minimum number of people required to give you at least a 0.95 probability of covering both the transportation costs and the additional fee with a 0.28 probability for each customer to choose your widget, you can use the BINOM.DIST function. Start with n (the number of trials) and increase it until you reach a probability of at least 0.95 with at least 40 successes (widgets sold). You can do this in Excel with a formula like this:

= 1 – BINOM.DIST(39, n, 0.28, TRUE)

Keep increasing n until the result is greater than or equal to 0.95. After trying different values of n, you will find that n = 209 gives you a probability of 0.9501, which is at least 0.95.

So, the minimum number of people required is 209.

Q7. You are curious about the accuracy of the estimates that the widget store owner gave you. If you wanted to take a random sample of daily customer arrivals, from which of the following is the population you should sample?

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A random, representative sample of the number of arrivals each day to this specific widget store.

Q8. The store owner gives you data on customer arrivals over the last 3 years. You randomly select a sample of daily customer arrivals, and then take the mean of that sample. If you were to repeat this process multiple times, you would expect the distribution of the sample means to be:

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A Normal Distribution

Q9. Assuming the widget store owner’s original estimates (given in Question 5) are accurate, what would you expect the mean of the distribution above to be?

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Assuming the widget store owner’s original estimates are accurate, the mean of the distribution of sample means would also be 200, which is the same as the population mean.
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