Probability distributions

Probability Distributions in plane words

A probability distribution is an illustration of the probabilities of different values for a given variable.

Example


Let us assume that our variable is the outcome of a die roll. Each of the 6 possible outcomes of the die roll is equally probable, thus the probability of any one outcome is 1/6. The probability distribution for the variable will assume a box-like shape, as shown below. The distribution illustrates that all the different values of the die are equally probable.

When we work with data, it is rare that all outcomes are equally probable, as with the roll of a die. Usually, there will be many values that are about average, which have the highest probability of occurring.

Imagine that we weigh the students in your class, and group our observations. The probability distribution could, for example, look like this:

The average weight is obviously in the middle. Most observations are close to the average, with individual students weighing more or less than the average.

When we work with large amounts of data, distributions give us a quick overview of the most probable values. If we, for instance, develop a similar analysis of weight distribution for all 5 million Danes, we could use a breakdown like this to illustrate the proportion of the population who are in different weight ranges.

Relative Frequency and Theoretical Distributions

Just as distributions can be based on the data you collect, distributions can also be based on theoretical probabilities.

To explain the difference, let us assume that you want to see the height distribution for all Danes over the age of 18.

Since you do not have time to measure the height of all Danes, we instead take a sample of 100 people. Then, you group the observed height values into ranges and calculate the relative size of the ranges, as shown below:

This is called a relative frequency distribution. The bar graph shows the proportion of observations in each interval relative to the total number of observations. To put it another way, it describes the distribution of the relative probability of the different observations in the sample.

A theoretical probability distribution is based on a mathematical presentation of the situation. A theoretical probability distribution can be seen as a model that reflects the theoretical probabilities for an entire population, such as how the height of the adult population is distributed.

Theoretical probability distributions are universal. They can be used for all variables, as long as the variables meet certain criteria. Often, we don’t have enough resources available to collect data for an entire population. In this context, a theoretical probability distribution is helpful as a model for how data are distributed throughout the population.

Different Distributions

There are many different distributions. What distinguishes them is a number of assumptions which we must look into. The fundamental distinction between distributions is based on whether they use discrete or continuous variables.

As we said in the section on random variables, a discrete variable is limited to a single unit, such as a car or a house.

Continuous variables are the opposite of discrete variables. They do not have units that can be accurately counted, only measured. Examples of continuous variables are temperature, weight and speed. With continuous variables, the probability of a given value appearing is zero.

Specifically, the probability that a June day will be exactly 20 degrees is zero. We may reason that 20 degrees could just as well be 19.99999 or 20.000001. Theoretically, 20 degrees is a value that could assume an infinite number of decimal places and, therefore, cannot be measured exactly.

As mentioned, there are different types of distributions. We will look at the most common, which can be classified as continuous or discrete.

To provide an overview of the differences in individual distributions and how they are used, we will briefly review the properties of individual distributions. Then, we will look at a case study where the distributions are used in a practical context.

Binomial distribution

Model: X~b(n, p)

Parameter: Where "n" is the sample size and "p" is the marked portion of the population.

Assumptions:

  • Discrete variable—the variable can be measured in whole units.
  • “n” experiments observed. Each observation in the sample is understood to be the out-come of one of “n” number of experiments.
  • Constant probability. The outcomes of individual experiments are independent of one another.
  • Each element can be defined either as marked or unmarked. The term binomial designation refers to two possible outcomes.

Formulas:

Where “p” is marked portion of the population and “n” is the sample size and where

Mean and variance:

E (X) = n . p VAR(X) = n . p . (1 - p)

Example


A trustee knows from experience that there is a 15% probability that she will choose a share that produces a negative return after one year. She has just invested in 10 different shares and wants to calculate the probability that all shares will be profitable after a year.

​Solution

X: The share provides a negative return after a year

X ~ b(p = 0,15, n = 10)

Formula Calculation:

Calculation using the Statlearn program:

P(X = 0) = 0,19687

Thus, there is a 19.69% probability that there will be zero shares that give a negative return after a year. Conversely, there is about an 80% probability that at least one of the shares will provide a negative return

Hypergeometric Distribution

Model: X~h(N, m, n)

Parameters: N = population size, m = number of marked items in the population and n = sample size.

Assumptions:

  • Discrete variable (the variable can be measured in whole units).
  • “n” elements taken from a finite population “N” (N must be known, or at least be possible to calculate/count).
  • Continuous probability does not apply. The outcome of individual experiments needs to be interdependent.
  • Each element can be defined as marked (m) or unmarked (m).

Formulas:

Mean and variance:

If p is calculated as m/n then the following applies:

Example


In a particular version of the lottery, there are 12 balls numbered from 1-12. Balls 1-4 are considered winning balls

Suppose that 4 balls are drawn randomly from among the 12. What is the probability that all the balls drawn will be winners?

Solutions:

X: A winning ball is drawn (ball 1 2 4)

X~h(N = 12, m = 4, n = 4)​

Formula calculation:

Calculation using the Statlearn program:

  • Select the distribution.
  • Select the hypergeometric distribution and paste the values.

P(X = 4) 5 0,00202

Thus, the probability of being lucky enough to win with all 4 balls is 0.2% (2,000)


Poisson Distribution

Model: X ~ Ps(λ)

Parameters:

Intensity, λ, can be interpreted as the average number of instances in a given time or within a specified quantity.

n 5 number of periods as based on λ.

Assumptions:

  • Discrete variable (the variable must be measured in whole units)
  • Number of instances of marked elements observed over time or measured per quantity, such as the number of accidents per month or the number of female students per class.
  • Elements occur independently in the given period (i.e., constant probability, as in a binomial distribution).

Formulas:

Where λ represents the population intensity and e is a constant with a value of 2.718.

Mean and variance:

Example


On average, 3.4 patients per day visit a certain hospital with a broken leg. What is the probability that, on a random day, this hospital will receive 5 patients with a broken leg?

Solutions:

X: Patient with a broken leg.

X~Ps(λ = 3,4)

Formula Calculations:

Calculation using the Statlearn program:

  • Select distribution.
  • Select the binomial distribution and paste the values.

Interpretation:

There is a 12.6% probability that, on any given day, 5 patients will come to the hospital with a broken leg.

Parameters:

Assumption:

  • Data is normally distributed.
  • Continuous variable. If the variable were discrete, it could be approximated, i.e., placed in a normal distribution approximation, if the conditions were met. A more detailed explanation of approximation conditions can be found in Appendix 1 at the end of this chapter.

Formula:

All normally distributed variables are transformed into Z-score with the above formula. A z-score is the number of standard deviations an observation is from the center of a probability.

A normal distribution is a cumulative density function7 (f) and is used when working with continuous variables. Continuous variables are, in contrast to discrete variables, not limited to one complete unit.

Therefore, a normal distribution corresponds to the probability of an observation falling into one of the intervals given, not to the probability of getting any exact value.

For example, if we were to calculate the probability that the temperature would get between 24 and 25 degrees, there would be no chance that it would be exactly 25 degrees. For this same reason, we use normal distribution symbols, most often “≤”, and especially “≥” but not “=”.

Example


A meteorologist found that, from June 12 to June 16, the average temperature was 19.3 Celsius, with a standard deviation of 4.5 degrees

What is the probability that it will be at least 22 degrees Celsius?

Solutions:

X: Temperature during the daytime (in Celsius) in June.

X~N(µ = 19,3, σ = 4,5)

Calculation using the Statlearn program:

  • Select distribution.
  • Select the binomial distribution and paste the values.

P(X ≥ 22) 5 0,274

Thus, there is a 27.4% chance that it will be at least 22 degrees in the daytime in June.

List of Distributions and their Characteristics

Case Study

The company Gene Food specializes in selling foods with added plant extracts that have cholesterol-lowering properties. Since its launch in 2003, the company has experienced tremendous growth, but this growth has not been without its costs. A growing number of customers are starting to complain about the products. Customer complaints are aimed at the company’s three main products in particular:

  • Mango drinks sold in boxes of 30 bottles.
  • Chocolate bagels sold in bags of 2 kg.
  • Nuts sold in boxes of 500 grams.

In regard to the mango drinks, the complaint is that the labels are often upside down. With the chocolate bagels, the complaint is that they sometimes don’t have any chocolate coating. When it comes to the nuts, the customers say that the boxes sometimes weigh significantly less than the indicated 500 grams.

As you may have guessed, all three quality problems are related to probability.

In the case of the mango drinks, there is the probability that the machine will apply the label upside down. With the chocolate bagels, there is the probability that the coating machine won’t apply enough chocolate. There is also the probability that the machine adding the nuts will not fill the packages up. As you can see, the conditions vary for each of the quality problems.

Regarding the bottles, the variable is discrete. We can reasonably assume that the probability that the labeling machine will make an error is constant. If the labeling machine applies a label upside down on a bottle, that does not affect the probability that it will apply the label upside down on the next bottle

With chocolate bagels, the variable is also discrete. However, while there are exactly 30 bottles in a box of mango drinks, the size and weight of each bagel varies, so we do not know exactly how many bagels will be in each bag.

The last complaint is that the boxes of nuts often weigh less than 500 grams. The amount of nuts is measured as part of the weight of the boxes and is, therefore, a continuous variable.

To get an overview of current production issues, you collect information from your experienced production manager, who tells you the following:

  • The probability of a bottle labeling error is 5%.
  • There are, on average, 0.4 bagels that do not get coated with chocolate.
  • The machine that fills the boxes of nuts produces an average of 505 grams of nuts in each box, with a standard deviation of 15 grams.

In the case of the mango drinks, we must measure the number of bottles with labeling errors. The number of bottles can be measured in full units and is thus a discrete variable.

The challenge now is finding the right discrete distribution. Since we are working with a speci-fic number of experiments and not an average or time interval, we can disregard the Poisson distribution.

So, the question is whether the variable follows a hypergeometric or a binomial distribution. For a binominal distribution to be used, the probability must be constant, i.e., the outcome of events and errors should not affect one another.

If we assume that giving a bottle an upside down label does not affect the probability that the next bottle will also receive an upside down label, the probability is constant. This indicates a binomial distribution.

From the production manager, we know that there is a 5% probability that a bottle will be labeled incorrectly. So, what is the probability that an entire box of 30 bottles will not include any labeling mistakes?

Steps for using a binomial distribution:

Variable: X: Bottle with labeling error

Model: X ~ b(p = 0, 05, n = 30)

Where “p” is the probability of an outcome of the variable (labeling error) and “n” is the sample size.

Assumptions

  • Discrete variable (whole bottles).
  • “n” elements observed (30 bottles per box).
  • Constant probability, i.e., independence between the individual experiments.
  • Each element can be defined as marked or unmarked.

Calculation using the Statlearn program:

  • Select distribution.
  • Select the binomial distribution and paste the values.

P(X = 0) = 0,215

There is a 21.5% probability that a box will contain only bottles without labeling errors. It is somewhat worrisome that the probability of a customer receiving some bottles with labeling errors whenever they buy a box of mango drinks is approximately 80%.

To ensure that your customers do not receive defective goods, you and your staff put a lot of effort into checking the goods at the warehouse. Among 100 pallets of boxes of mango drinks, you have found 7 pallets that need to be discarded.

By mistake, the 7 defective pallets were not discarded, but moved back with the other pallets. Your biggest customer has just ordered 15 pallets of the mango drinks and you, therefore, want to quickly calculate the probability that he has received one or more pallets of defective goods.

Just like bottles, pallets are a discrete variable. They can be measured in whole units. However, unlike in the case of the bottles, we know exactly how many items “m” (defective pallets = 7) there are in our population “N” (all pallets = 100).

We can, therefore, know the probability of getting a defective pallet:

Depending on whether you get a normal or a defective pallet, the probability of getting another defective pallet changes:

When we have a discrete variable (a pallet) and a known population (100 pallets) with a certain number of marked items (defective pallets), we have a hypergeometric variable.

Thus, we can determine the probability that, among the 15 pallets that the customer ordered, there will be one or more defective pallets.

Steps for using a hypergeometric distribution:

Definition of variable: X: Defective pallet

Model: X ~ h(N = 100, m = 7, n = 15)

Where “N” is the number of elements in the population, “m” is the number of marked items, and “n” is the sample size. Assumptions:

  • Discrete variable (one pallet).
  • “n” elements taken from finite population “N” (a customer buys 15 pallets out of a total of 100 pallets).
  • Probability is not constant, i.e., the individual experiments are dependent on one another.
  • Each element can be defined as marked or unmarked (non-defective pallet/defective pallet).

Calculation using the Statlearn program:

  • Select distribution.
  • Select the binomial distribution and paste the values.

P(X ≥ 1) 5 0,6916

Thus, there is a 69% probability that there will be at least one defective pallet among the 15 the customer has ordered.

As far as the quality problems with the chocolate bagels go, customers complain that bagels often lack a chocolate coating. The production supervisor has estimated that, on average, 0.4 bagels out of each 2 kg bag lack a chocolate coating. In light of this information, you want to know the probability that a given bag will contain one or more bagels without a chocolate coating.

Since the size and weight of the bagels varies, each 2 kg bag does not always have the same number of bagels. Instead of working with a precise number of experiments, as in the mango drinks and defective pallets examples, we need to use an average. However, like the two variables mentioned before, a bagel is a discrete variable, which leads us to use the Poisson distribution.

What is the probability that a single bag includes some bagels without a chocolate coating?

Steps for using a Poisson distribution:

Definition of variable: X: “Chocolate” bagel without chocolate coating

Model: X ~ Ps(λ = 0,4)

Assumptions:

  • Discrete variable (whole bagel).
  • Number of instances of marked elements observed in a specific quantity (2 kg bag).
  • Outcomes occur independently. When a bagel does not receive the right amount of coa- ting, this does not change the probability that the next bagel will also have this error. The errors are independent from one another.

Calculations using the Statlearn program:

  • Select distribution.
  • Select the Poisson distribution and insert the values.

Interpretation:

There is a 67% probability that a bag of bagels will not contain defects, i.e., bagels without a chocolate coating. Conversely, there is a 33% chance that a bag will contain defective bagels, which is not quite satisfactory for customers.

Steps for using a normal distribution:

With the boxes of nuts, customers complain that the boxes often weigh less than the 500 grams listed on the package. In the previous examples, we could count the exact number of labeling mistakes, defective pallets, and uncoated “chocolate bagels”.

However, with the nuts, we are dea- ling with the weight per box. Weight must not be treated like a discrete variable. When we weigh an object, we do not know the exact weight, as this could be specified with an infinite number of decimal places. For this reason, weight is classified as a continuous variable. This leads us to use a normal distribution.

The normal distribution is the mother of all distributions. It plays a central role in statistics. The term normal distribution is used because, as you have guessed, most variables follow a normal distribution. They have a bell-shaped probability distribution with the average in the middle and an equal number of observations on each side

To understand the idea of a normal distribution, let us suppose that we take a sample of 10 boxes of nuts and weigh each box. Subsequently, we divide the observations into categories, as shown below:

The distribution is clearly asymmetric. It has little in common with the normal distribution you have just seen. But, if we increase the sample size to 100 boxes, you will see a much more symmetrical (normal) distribution.

The reason for this transformation is that we have more observations covering a wider range of values. This means that we can divide the observations into smaller and narrower ranges, which helps smooth out the density function curve.

As we continue to increase the sample size, our intervals will become infinitely small, which will make our distribution resemble a smooth curve with the average in the middle and nearly half of the observations on each side of the average. A variable that fits the above description is normally distributed.

Normally distributed variables are actually variables that fit a standard normal distribution. The normal distribution is the mathematical model underlying the example of the boxed nuts we have just discussed.

Standard normal distributions are based on so-called z-values, which correspond to the number of standard deviations from the mean, i.e., the middle of the distribution. The standard normal distribution is characterized by a fixed relationship between the number of standard deviations (z-values) from the average and the actual area of distribution, as illustrated in the figure below.

For example, we can see that there is a 68% probability that a random observation will be in the range of ±1 standard deviation from the mean when the variable follows a normal distribution.

Returning to the example of the boxed nuts, you know that the machine places an average of 505 grams of nuts in each box, with a standard deviation of 15 grams.

Since weight is a continuous variable, it is assumed that the variable will be normally distributed8. That means about 68% of our observations (boxes) will be in the range of 490 to 520 grams. It is now clear that a significant proportion of boxes will weigh less than the weight indicated on the boxes.

What is the probability that a box will weigh more than 500 grams?

Watch the video explanation of the normal distribution

Steps for using using a normal distribution:

Definition of variable: X: The weight of a box of nuts

Model: N(µ = 505 grams, σ = 15 grams)

Assumptions:

  • Data is normally distributed.
  • Independence between the individual observations.

Calculations using the Statlearn program:

  • Select distribution.
  • Select the Poisson distribution and insert the values.

P(X ≤ 500) 5 0,3694

Interpretation:

Thus, there is a 36.94% probability that a box will weigh more than 500 grams when the average is 505 grams, with a standard deviation of 15 grams. Therefore, about one third of the boxes will be overfilled.

Summary

We have now touched on four basic distributions, three discrete and one continuous.

A binomial distribution is used when we have a continuous probability, i.e., independence between individual experiments. In the example of the mango drinks, there was a continuous probability that the label machine would make a mistake.

There was, in other words, independence between each bottle passing through the labeling machine. Unlike with a binomial distribution, probability is not constant with a hypergeometric distribution.

In a hypergeometric distribution, the probability of each marked item changes every time a sample is taken. This is because the population contains an exact number of marked items. In the lottery example, there were 4 winning balls (marked items) out of a total of 12 balls (population).

Every time a ball was taken from the population (all lottery balls), this changed the probability of selecting the remaining winning balls, because after each draw, there were a smaller number of balls remaining.

A Poisson distribution, like the previous two distributions, is based on discrete variables. In contrast to a binomial distribution, where there are a fixed number of experiments, we do not have this information for variables that follow a Poisson distribution.

In the case of the chocolate bagels, we did not know exactly how many bagels were in each bag. Therefore, our starting point was the average number of defects per bag of bagels.

A normal distribution is used when working with continuous variables. In the example involving the boxes of nuts, the variable was continuous, since weight is a measurable unit that can be specified with an infinite number of decimal places.

As we mentioned, all normally distributed variables fol- low the standard normal distribution, which is based on a fixed relationship between the number of standard deviations from the average and the area of distribution. The number of standard deviations from the mean is expressed in a standard normal distribution using z-values.

Probability Distribution Exercises

The first problem set tests your comprehesion of the terms covered in this chapter. In your own words, try to explain the importance of the different distributions discussed in this chapter. The second problem set requires you to make calculations.

Interpretation Exercises

  • Why use probability distributions in general?
  • How would you explain the difference between a discrete and a continuous probability variable?
  • How would you explain the difference between a binomial and hypergeometric variable?
  • In what contexts should we use the Poisson distribution?
  • With both the Poisson and normal distribution, an average is used to calculate probabilities. How do you know whether you should use the Poisson distribution instead of the normal distribution?

Calculation Exercises

Problem 1

The three questions below provide an exercise in finding the right distributions. You only need to decide which distribution works for each question. For all three questions, it is important to try to define the variable (X) and use the correct notation for the different distributions.

Question 1. At the annual meeting of the UN, top officials gathered from all nations. What is the probability that a random sample of 10 officials will include three from Africa?

Question 2. We know the given probability of an accident occurring on a particular stretch of a highway named Lyngby. What are the chances that 4 accidents will occur per 2,000 vehicles are driven over this route?

Question 3. A realtor knows that, on average, 10 homes are sold per month during the summer months. What is the probability that 30 homes will be sold during a two- month period in the summer?

Problem 2

Suppose that there is a 40% probability that a student at Niels Brock enrolls directly after high school. We are working with a sample of 100 students.

Question 1. Which distribution works with our variable? (Please elaborate.)

Question 2. What is the probability that the sample contains more than 40, but no more than 70, students who came directly from high school?

Problem 3

A group of travelers are arriving in Barbados on two different planes: A and B. Group A is composed of a total of 100 tourists, 30 of whom traveled on plane A. You, the tour guide, have been informed that 5 of the 100 suitcases have been lost. You must now notify group A, who just landed.

Before you speak to Group A, what is the probability that any of the lost suitcases belonged to this group?

Question 1. Define the distribution and explain your choice.

Question 2. What is the probability that no one in Group A has lost any luggage? (Remember to define the variable).

Question 3. What is the probability that only one person from Group B has lost a suitcase? (Remember to define the variable.)

Problem 4

A doctor needs to order a new shipment of EPO9 for patients whose ability to take in oxygen is impaired. The doctor knows that there are 100 patients who take 10 ml of EPO on a daily basis and that the drug works in 90% of patients. The remaining 10% of patients need a double dose of the drug. With a double dose, it is assumed that EPO has a 100% probability of working.

Question 1. What is the probability that, on any given day, a single dose of EPO will work for less than 83% of patients?

Question 2. What is the probability that at least 9 and no more than 12 patients will need a double dose on any given day?

Question 3. What is the probability that, during any given week (7 day period), fewer than 65 patients will need to be given a double dose?

Question 4. How many liters of EPO should the physician expect to be used up during the course of a week?

Problem 5

A stockbroker has been following Nokia’s share price for a long time. She has observed that the share price increases by an average of 5% when the company’s quarterly account earnings increase beyond the expected amount, with a standard deviation of 1.3%.

Assume that the next quarter’s results will be better than expected.

What is probability that the share price will increase by at least 6.5%?

Problem 6

The Sales Director of BMW Odense expects that, during the summer months, an average of 30 cars will be sold per month.

Question 1. The manager needs your help to adjust inventory levels. What is the probability that between 30 and 40 cars will be sold during the month of June?

Question 2. What is the probability that over 200 cars will be sold during the summer?

Problem 7

Let us assume that the small bottles of Coca-Cola served on airplanes are about 15 ml. in volume. Assume they are filled to a height of 10 cm, with a standard deviation of 1 cm. If they are filled to a height of 8.2 cm or less, they are considered underfilled. If they are filled to a height of over 11.5, they are considered overfilled.

Question 1. What is the probability that a bottle will be overfilled? Illustrate the corresponding area with a normal distribution.

Question 2. What is the probability that a bottle will either be over or underfilled? Illustrate as in question 1.

Question 3. If a box contains 30 Coca-Cola bottles, what is the probability that the average fill height for a whole box will exceed 8.2 cm?

Question 4. If we have a z-value of 3, how does that correspond to the filling height of a single bottle? What is the filling height if the z value is -1?

Problem 8

Suppose that the number of complaints received by Nordea Customer Service each month follows a Poisson distribution, with an average of 255 complaints.

Question 1. What is the probability that Nordea:

  • Will receive more than 300 complaints in a given month?
  • Will receive a maximum of 750 complaints in 3 months?
  • Will receive no complaints for one entire day (assuming 23 working days/month)?

Question 2. What is the probability that Nordea will receive exactly 255 complaints each month for two months in a row? Assume independence between the number of monthly complaints.

(Tip: Use the Intersection.)

Solutions for Probability Distributions Problems

The following page show solutions for the exercises involving probability distributions. For the sake of simplicity, we are just showing the results. You will need to get the other information on your own. For an exam, it is important that you show your work, including the details provided by the calculations that the Statlearn program performs.

Interpretation Exercises

  • Distributions are used to calculate probabilities in an efficient manner. The strength of distributions comes from the accumulation of probabilities. Without distributions, it would be necessary to manually calculate the probabilities for each value of a random variable. For example, the likelihood that P(X ≤ 3) can be calculated by summing up the probabilities for four separate calculations: P(X = 0) 1 P(X = 1) 1 P(X = 2) 1 P(X = 3). However, with a probability distribution, P(X ≤ 3) can be read directly in the distribution, thereby avoiding time-consuming calculations.
  • A discrete variable can be measured in whole units, such as a house or a car. Continuous variables, on the other hand, cannot be limited to whole units. Continuous variables are characterized by the fact that they cannot be measured exactly. The value of a continuous variable could theoretically be measured to an infinite number of decimal places. Examples include temperature, velocity and time.
  • A binomial variable is characterized by a constant probability for each outcome of an experiment. When rolling a die, the probability of any given outcome is 1/6. No matter how many times the die is rolled, there remains a probability of 1/6 for a particular outcome— hence the constant probability. A hypergeometric variable represents the opposite situation. In a lotto with 36 balls, there is a probability of 1/36 of you removing a certain ball at the first draw. After each draw, the probability changes, because there are fewer balls remaining.
  • The Poisson distribution is used for discrete variables, when events occur independently of each other in a given time interval.
  • The Poisson distribution is meant to be used with discrete variables, whereas a normal distribution should be used with continuous variables.

Calculation Exercises

Problem 1

Question 1.

X: Number of UN officials from Africa.

X ~ h(N = all UN officials, m = UN officials from Africa, n = 10). P(X = 3)

Assumptions:

  • Discrete variable.
  • n elements taken from a finite population (N).
  • Probability is not constant, i.e., there exists dependence between the individual experiments.
  • Each element can be defined as marked or unmarked (African, non-African).

Question 2.

X: Number of accidents on the Lyngby highway. 

X ~ b(p, n)

P(X = 4)

Assumptions:

Question 3.

X: Number of homes sold per month during the summer. 

X ~ Ps(λ = 20)

P(X = 30)

Assumptions:

  • Discrete variable.
  • Number of marked elements observed over time (sold houses).
  • The elements are independent.
  • The probability of a marked element appearing is constant.

Problem 2

Question 1.

X: Number of Niels Brock students who come directly from high school.

X ~ b(p = 0,4 n = 100) since the probability of a student coming directly from high school is constant.

Question 2.

P(41 ≤ X ≤ 70) = 0,456706

Problem 3

Question 1.

X ~ h(N = 100 m = 5 “n” is not yet known)

The hypergeometric distribution should be used because the likelihood of a lost suitcase belonging to a tourist changes every time a tourist is selected. Thus, the probability of a lost suitcase belonging to a tourist is not constant.

Question 2.

X: Number of tourists who lost their suitcases from group A.

X ~ h(N 5=100, m = 5, n = 30)

P(X = 0) = 0,160757

Question 3.

Y: Number of tourists who lost their suitcases from Group B. 

Y ~ h(N = 100, m = 5, n = 70)

Problem 4

Question 1.

X: Number of patients who only need a single dose of EPO.

X ~ b(p = 0,9, n = 100) when the variable is discrete and there is independence between patients who need EPO.

P(X ≤ 82) 5 0,010007

Question 2.

Y: Number of patients who need a double dose of EPO. 

P(9 ≤ X ≤ 12) 5 0,4809

Question 3.

Y: Number of patients who need a double dose of EPO in a week (7 days). 

Y ~ b(p = 0,1 n = 700) P(X ≤ 64) = 0,246935

Question 4.

X: Single dose of Epo (10 ml). 

Y: Double dose of Epo (20 ml).

E(X) = n * p = 700 * 0,9 = 630 → 10 ml * 630 = 6300 ml

E(Y) = n * p = 700 * 0,1 = 70 → 20 ml * 70 = 1400 ml

E(X) 1 E(Y) = 6300 + 1400 = 7700 ml = 7,7 l

Problem 5

X: Nokia’s share price when the quarterly financial account earnings have increased beyond what was expected. 

X ~ N(µ = 0,05 σ = 0,013)

The normal distribution is selected, since the share price is a continuous variable. 

P(X ≥ 0,065) = 0,124282

Problem 6

Question 1.

X: Number of cars sold per month in summer (6 months). 

X ~ Ps(λ = 30)

P(30 ≤ X ≤ 40) = 0,491973

Question 2.

X: Number of cars sold in the summer. 

X ~ Ps(λ = 180)

P(X≤ 201) = 0,065206

Problem 7

Question 1.

X: Fill height of a single bottle (in cm).

X ~ N(µ = 10 , σ = 1) since height is a continuous variable. 

P(X ≥ 11,5) = 0,0668

Question 2.

P(X ≤ 8,2) + P(X ≥ 11,5) 5 0,0359 + 0,0668 = 0,1027

Question 3.

X: Average fill height for a box of 30 bottles.

Question 4.

An average fill height of 10 cm with a standard deviation of 1 cm corresponds to a z-value of 3 for a fill height of 13 cm and a z-value of -1 to a fill height of 9 cm. Z-values correspond to the number of standard deviations from the center of the distribution (average).

Problem 8

Question 1.

  • P(X ≥ 301) = 0,002714
  • P(X ≤ 750) = 0,301576
  • P(X =0) = 0,000017 this is based on an average of 11 (255/23).

Question 2.

P(255 complaints per month 1 ∩ 255 complaints per month 2) = 0,024975 * 0,024975 = 0,000624

Appendix 1: Approximation Rules

Several prerequisites must be met before you can try to approximate to a normal distribution:

From the hypergeometric to the normal distribution:

From binomial to normal distribution:

n * p̂ (1 - p̂) * 9 if met → Approximation OK 

From Poisson to normal distribution:

n * λˆ > 9; if met → Approximation OK

The Statlearn program automatically tests the confidence interval hasn’t been introduced yet whether approximation conditions are met.