Understanding Probability Distribution, Variance, and Standard Deviation in Experimental and Theoretical Contexts

There are a variety of experiments whose outcomes can be reduced to success or failure. This lesson will explore how real-life situations that satisfy specific conditions can be modeled as binomial experiments. Additionally, methods of determining the probability of a certain number of successes in

n

binomial trials will be presented.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.
Understanding Probability

Theoretical Probability
Experimental Probability
Random Variable

Understanding Descriptive Measures

Expected Value of a Random Variable
Standard Deviation

Understanding Types of Data

Continuous Quantity
Discrete Quantity
Examining Decisions

Other Recommended Readings

Pascals Triangle
Binomial Theorem

Explore

Investigating Distributions Using Experiments

The Galton board is a device patented by Sir Francis Galton. It consists of a set of balls that are dropped from the top of the board. As the balls fall, they move to the left or right every time they bounce off of the pegs embedded in the board until they land in one of the bins at the bottom. Each path option has the same probability.

A Galton Board with a set of marbles ready to be dropped.

Based on the outcomes, try to ask the following questions.

Which bins would be expected to collect the most balls?
About how many balls are expected to land in each bin?
How many paths branch out from each peg?

Discussion

Analyzing the Probabilities of the Values of a Random Variable

A random variable assigns a numerical value to an outcome of a probability experiment. In many situations, it is important to know how likely it is that a random variable will take a specific value. This can be represented by listing or graphing the probability of each value of a random variable. This is called a probability distribution.

Discussion

Can the Experimental Distribution Estimate the Theoretical Distribution?

The expected value of a random variable $X$ is the average of the possible outcomes of a random variable. It is used to describe the center of a probability distribution. For a discrete random variable, the expected value $E (X)$ is given by the weighted mean.

$E (X) = i = 1 \sum n x_{i} \cdot P (X = x_{i})$

In this formula, $x_{i}$ represents a specific outcome, $P (X = x_{i})$ corresponds to the associated probability of $x_{i},$ and $n$ is the number of all possible outcomes. According to the law of large numbers, when considering a sequence of random variables, its average tends to the expected value under specific conditions.

Discussion

How is the Variation of a Probability Distribution Measured?

The expected value is commonly used with a measure of variation such as the variance or standard deviation to determine how outcome will differ from the expected value.

Concept

Variance of a Random Variable

The variance of a random variable describes how far from the expected value

E (X)

the outcomes of a random variable

X

are likely to be. To calculate the variance, begin by determining the deviation of each possible outcome

x_{i}

— the difference between

x_{i}

and

E (X) .

\underline{Deviation of x_{i}} x_{i} - E (X)

The variance is the total sum of the products of the deviation of each outcome and its corresponding probability

P (X = x_{i}) .

σ^{2} = Σ [[x_{i} - E (X)]^{2} \cdot P (X = x_{i})]

The variance is denoted by

σ^{2}

because it is the square of the standard deviation

σ .

Concept

Standard Deviation of a Random Variable

The standard deviation of a random variable is a measure of variation that describes how spread out the outcomes of a random variable $X$ are from its expected value $E (X) .$ The standard deviation is represented by the Greek letter $σ$ — read as sigma — and is given by the square root of the variance of $X .$

σ = Σ [[x_{i} - E (X)]^{2} \cdot P (X = x_{i})]

In this formula, $x_{i}$ is a specific outcome and $P (X = x_{i})$ is the probability of $x_{i} .$

Example

Let $X$ be the random variable representing the number of cars sold on a given day in a car dealership. The table below shows the probability distribution of $X .$

$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$\frac{1}{1 5}$	$\frac{3}{1 5}$	$\frac{6}{1 5}$	$\frac{3}{1 5}$	$\frac{2}{1 5}$

Use the probability distribution to calculate the standard deviation of

X .

Begin by finding the expected value.

E (X) = [x_{1} \cdot P (X = x_{1})] + [x_{2} \cdot P (X = x_{2})] + \dots + [x_{n} \cdot P (X = x_{n})]

SubstituteValues

Substitute values

E (X) = 0 (\frac{1}{1 5}) + 1 (\frac{3}{1 5}) + 2 (\frac{6}{1 5}) + 3 (\frac{3}{1 5}) + 4 (\frac{2}{1 5})

Evaluate right-hand side

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

E (X) = \frac{0}{1 5} + \frac{3}{1 5} + \frac{1 2}{1 5} + \frac{9}{1 5} + \frac{8}{1 5}

$\frac{0}{a} = 0$

E (X) = 0 + \frac{3}{1 5} + \frac{1 2}{1 5} + \frac{9}{1 5} + \frac{8}{1 5}

IdPropAdd

\IdPropAdd

E (X) = \frac{3}{1 5} + \frac{1 2}{1 5} + \frac{9}{1 5} + \frac{8}{1 5}

AddFrac

Add fractions

E (X) = \frac{3 2}{1 5}

UseCalc

Use a calculator

E (X) = 2.1333 \dots

RoundDec

Round to $2$ decimal place(s)

E (X) \approx 2.13

Next, the variance of

X

will be calculated using a table of values. To do so, calculate the square of each deviation — the difference between each possible value of

X

and

E (X) .

Then multiply the square deviation of each value by its corresponding probability.

$x_{i}$	$[x_{i} - E (X)]^{2}$	$[x_{i} - E (X)]^{2} \cdot P (X = x_{i})$
$0$	$(0 - 2.13)^{2} = 4.5369$	$4.5364 \cdot \frac{1}{1 5} \approx 0.3025$
$1$	$(1 - 2.13)^{2} = 1.2769$	$1.2769 \cdot \frac{3}{1 5} \approx 0.2554$
$2$	$(2 - 2.13)^{2} = 0.0169$	$0.0169 \cdot \frac{6}{1 5} \approx 0.0068$
$3$	$(3 - 2.13)^{2} = 0.7569$	$0.7569 \cdot \frac{3}{1 5} \approx 0.1514$
$4$	$(4 - 2.13)^{2} = 3.4969$	$3.4969 \cdot \frac{2}{1 5} \approx 0.4663$
Variance $σ^{2}$		$\approx 1.1824$

Finally, calculate the square root of the variance to get the standard deviation of $X .$

\underline{Standard Deviation :} 1.1824 \approx 1.09

Variance vs Standard Deviation

The standard deviation is preferred over the variance because taking the square root of the variance gives the standard deviation the same units as

X .

This makes it possible to compare the possible outcomes relative to the expected value.

Example

Using Standard Deviation to Make a Decision

Izabella's aunt Magdalena owns a clothing store. She needs to increase stock in her shop and plans to invest $$ 15000$ in one of the two collections that were offered to her by well-known brands. Each brand claims that they have a great expected rate of return. Their probability distributions are described below.

Dylan and Izabella want to help Magdalena make the best decision. They decided to use their recently acquired knowledge about the expected value and standard deviation of a probability distribution to analyze the offers. Help them answer the following questions and give the best advice to Magdalena.

a Pair each description with its corresponding measure.

b Which investment should Izabella and Dylan advise Magdalena to choose?

Hint

a The expected value is the total sum of the products of every possible value of the random variable and its corresponding probability.

b Compare the expected values and the standard deviations of the distributions.

Solution

a To match each description with its corresponding value, find the expected value and the standard deviation of each probability distribution one at a time.

Collection I

The expected value

E (X)

of a discrete random variable is given by the total sum of the products of every possible value

x_{i}

and its associated probability

P (X = x_{i}) .

Let

X

represent the profit of the first collection. Note that each loss will be represented by a negative value.

E (X) = [x_{1} \cdot P (X = x_{1})] + [x_{2} \cdot P (X = x_{2})] + \dots + [x_{n} \cdot P (X = x_{n})]

SubstituteValues

Substitute values

E (X) = 1200 \cdot 0.5 + 1800 \cdot 0.2 + 900 \cdot 0.2 + (- 150) \cdot 0.1

Multiply

E (X) = 600 + 360 + 180 - 15

AddSubTerms

Add and subtract terms

E (X) = 1125

The expected value can now be used to find the standard deviation of the probability distribution. Consider the formula for the standard deviation.

σ = Σ [[x_{i} - E (X)]^{2} \cdot P (X = x_{i})]

To apply this formula, calculate the square of each deviation — the difference between each

x_{i}

value and the expected value — and then multiply it by its corresponding probability. This process will be shown in a table.

$x_{i}$	$[x_{i} - E (X)]^{2}$	$[x_{i} - E (X)]^{2} \cdot P (X = x_{i})$
$1200$	$(1200 - 1125)^{2} = 5625$	$5625 \cdot 0.5 = 2812.5$
$1800$	$(1800 - 1125)^{2} = 455625$	$455625 \cdot 0.2 = 91125$
$900$	$(900 - 1125)^{2} = 50625$	$50625 \cdot 0.2 = 10125$
$- 150$	$(- 150 - 1125)^{2} = 1625625$	$1625625 \cdot 0.1 = 162562.5$
Sum of Values		$266625$

The sum of the values of the last column represents the variance of the probability distribution. The standard deviation will be found by taking the square root of

266625 .

\underline{Standard Deviation :} 266625 \approx 516.36

Collection II

Follow a similar procedure as before to calculate the expected value for Collection II.

E (X) = [x_{1} \cdot P (X = x_{1})] + [x_{2} \cdot P (X = x_{2})] + \dots + [x_{n} \cdot P (X = x_{n})]

SubstituteValues

Substitute values

E (X) = 3600 \cdot 0.3 + 2850 \cdot 0.1 + (- 300) \cdot 0.4 + (- 500) \cdot 0.2

Multiply

E (X) = 1080 + 285 - 120 - 100

AddSubTerms

Add and subtract terms

E (X) = 1145

Now use a table to calculate the variance of the distribution.

$x_{i}$	$[x_{i} - E (X)]^{2}$	$[x_{i} - E (X)]^{2} \cdot P (X = x_{i})$
$3600$	$(3600 - 1145)^{2} = 6027025$	$6027025 \cdot 0.3 = 1808107.5$
$2850$	$(2850 - 1145)^{2} = 2907025$	$2907025 \cdot 0.1 = 290702.5$
$- 300$	$(- 300 - 1145)^{2} = 2088025$	$2088025 \cdot 0.4 = 835210$
$- 500$	$(- 500 - 1145)^{2} = 2706025$	$2706025 \cdot 0.2 = 541205$
Sum of Values		$3475225$

The square root of the variance will be calculated to find the standard deviation of the probability distribution of Collection II.

\underline{Standard Deviation :} 3475225 \approx 1864.20

Conclusion

The expected value and the standard distribution of each probability distribution have been calculated. The following table summarizes these measures.

Measures of the Probability Distributions
Expected Value of Collection I	$1125$
Expected Value of Collection II	$1145$
Standard Deviation of Collection I	$\approx 516.23$
Standard Deviation of Collection II	$\approx 1864.20$

b Consider the expected value of each distribution.

\underline{Expected Values} Collection I : Collection II : E (X) = 1125 E (X) = 1145

The expected values do not give much information on their own because they are very close. This means that the expected profit of each collection is similar. Next, compare the standard deviations to determine which distribution has more variability.

\underline{Standard Deviation} Collection I : Collection II : σ \approx 516.36 σ \approx 1864.20

The standard deviation of Collection II is almost four times the that of Collection I, which implies that the expected value of Collection II will have about four times the variability of Collection I. Since Collection II is riskier, with a high chance for both gains and losses, Izabella and Dylan should advise Magdalena to invest in Collection I.

Discussion

Binomial Experiments

The outcomes of many experiments can be reduced to two possibilities, success or failure. If two more conditions are satisfied, these experiments can be modeled by a binomial experiment.

Example

Classifying Whether Experiments Are Binomial

To better understand binomial experiments, Dylan and Izabella listed some examples of experiments. They want to determine whether the experiments can be modeled as binomial experiments. Which of these situations are examples of binomial experiments?

Hint

Is there a fixed number of trials for each experiment? How many outcomes are possible? Does the probability of each outcome remain constant for each trial? Are trials independent?

Solution

Start by recalling the conditions that a binomial experiment should satisfy.

There is a fixed number of independent trials.
Each trial has exactly two possible outcomes — success and failure.
For each trial, the probability of success is constant.

Analyze each situation one at a time to see if it follows all of the conditions of a binomial experiment.

The Scratch-off Cards

This situation has eight trials of selecting one scratch-off cards at random.

Each card could win a prize or not, which means there are two possible outcomes for each trial. Moreover, the probability of success, which is winning a prize, is

25 %

0.25,

for every card.

P (Success) = 0.25

Finally, the trials are independent because scratching one card does not affect the probability that any of the other cards reveals a prize. Therefore, this situation represents a binomial experiment.

Heights of $100$ People

Note that height can vary for every people surveyed.

Because there are $100$ possible answers, it is likely that more than $2$ different outcomes will occur. This means that this situation does not represent a binomial experiment.

Rolling a Die

In this case, rolling a

3

can take only one or many more trials.

It cannot be known how many rolls it will take until a

3

comes up. Therefore the number of trials is not fixed. This means that this experiment does not represent a binomial experiment.

Blood Type O

This situation has a fixed number of trials because it involves asking $20$ people if their blood type is O.

Each trial has only two possible solutions, blood type O or another type. Moreover, the probability of having blood type O in each trial is

0.4,

which represents the probability of success.

P (Success) = 0.4

Finally, because the answer of any person surveyed does not affect the probability that other people have O type blood, each trial is independent. This means that this situation is a binomial experiment.

Conclusion

All four situations have already been analyzed and the results are summarized in the following table.

Experiment	Is It a Binomial Experiment?
The Scratch-off Cards	$✓$
Heights of $100$ People	$\times$
Rolling a Die	$\times$
Blood Type O?	$✓$

Discussion

Probability of $x$ Successes in $n$ Binomial Trials

Because binomial experiments can simplify many complex situations, it is essential to determine how likely it is to obtain a specific number of successes out of $n$ trials in a given experiment. Also, the expected value, or center of the distribution, will be presented.

Rule

Binomial Probability

Let $X$ be the random variable representing the number of successes in $n$ binomial trials. The binomial probability $P (X = x)$ can be calculated by using the following formula.

$P (X = x) =_{n} C_{x} p^{x} q^{n - x}$

In this formula, $P (X = x)$ is the probability that the random variable $X$ is equal to $x,$ which means that there are exactly $x$ successes. Additionally, $p$ and $q$ are the probabilities of success and failure, respectively, and $_{n} C_{x}$ is the binomial coefficient.

Proof

This proof will begin by calculating the probability of obtaining a fixed sequence with $x$ successes in $n$ independent trials. The number of all possible sequences with $x$ successes will then be calculated. Finally, joining the first and second parts will give the formula.

The Probability for a Fixed Sequence

Note that there are

n

independent trials, of which

x

are successes. This means that the difference between

n

and

x

gives the number of failures.

Number of Succeses : Number of Failures : x n - x

Furthermore, let

p

be the probability of success and

q

be the probability of failure in one individual trial. Therefore, by the Multiplication Rule of Probability and the fact that trials are independent, the probability of

x

successes is given by multiplying

p

by itself

x

times.

Probability of x Successes : x times p \cdot p \cdot \dots \cdot p = p^{x}

Similarly, because there are

n - x

failures, the probability of

n - x

failures is given by multiplying

q

by itself

n - x

times.

Probability of n - x Failures : n - x times q \cdot q \cdot \dots \cdot q = q^{n - x}

Therefore, the probability of getting exactly

x

successes and

n - x

failures is given by the product of

p^{x}

and

q^{n - x}

p^{x} q^{n - x}

By the Commutative Property of Multiplication, any combination of

p

and

q

variables can be rearranged. Therefore, the expression is valid for any fixed sequence of

x

successes and

n - x

failures. However, note that this is the probability of only one of the possible sequences.

Finding the Total Number of Possible Sequences

Consider an experiment that consists of four independent trials. The sequences in which two successes

S

and two failures

F

occur are as follows.

{S S F F, S F S F, S F F S, - F S S F, F S F S, F F S S}

Note that as long as two successes occur, the order of the outcomes is not important. This can be simplified by considering how many ways the two successes can be arranged within the four trials since the remaining trials will automatically be failures. In such a situation, the combination formula can be used.

_{n} C_{r} = \frac{n !}{r ! ( n - r ) !}

SubstituteII

$n = 4$ , $r = 2$

_{4} C_{2} = \frac{4 !}{2 ! ( 4 - 2 ) !}

Evaluate right-hand side

SubTerm

Subtract term

_{4} C_{2} = \frac{4 !}{2 ! \cdot 2 !}

Write as a product

_{4} C_{2} = \frac{4 \cdot 3 \cdot 2 !}{2 ! \cdot 2 !}

CrossCommonFac

Cross out common factors

_{4} C_{2} = \frac{4 \cdot 3 \cdot 2 !}{2 ! \cdot 2 !}

CancelCommonFac

Cancel out common factors

_{4} C_{2} = \frac{4 \cdot 3}{2 !}

Multiply

_{4} C_{2} = \frac{1 2}{2 !}

$2! = 2$

_{4} C_{2} = \frac{1 2}{2}

CalcQuot

Calculate quotient

_{4} C_{2} = 6

This means that

6

different arrangements are possible, which corresponds to the number of sequences listed before. By following the same reasoning, different sequences of

n

trials with

x

successes can be found by calculating the number of combinations of

x

out of

n .

_{n} C_{x} = \frac{n !}{x ! ( n - x ) !}

Getting the Binomial Probability

The probability of obtaining a fixed sequence of

x

successes out of

n

trials is given by the product of

p

to the power of

x

and

q

to the power of

n - x .

p^{x} q^{n - x}

Moreover, there are

_{n} C_{x}

possible sequences with

x

successes. Therefore, by the Addition Rule of Probability, the probability

P (X = x)

of getting any of the possible sequences is given by adding the probability of getting one sequence

_{n} C_{x}

times.

P (X = x) =_{n} C_{x} times p^{x} q^{n - x} + \dots + p^{x} q^{n - x} ⇓ P (X = x) =_{n} C_{x} p^{x} q^{n - x}

Example

Finding the Probability of Guessing Correctly on a Multiple-Choice Test

Dylan is taking a test that consists of five multiple-choice questions. Each question has five answer choices, only one of which is correct. He did not have time to study, so he decides to guess every answer.

A paper with 5 questions and each question has 5 choices

What is the probability of guessing at least three answers correctly? Write the answer as a percentage. Round to two decimal places.

Hint

Consider using the Addition Rule of Probability.

Solution

To find the probability of guessing at least three answers correctly in a five question multiple-choice test, first, verify if this experiment satisfies the three conditions of a binomial experiment.

Condition	Given Experiment	Is Satisfied?
There is a fixed number of independent trials.	There are five trials — five questions.	$✓$
Each trial has two possible outcomes.	The answer for each question is either correct or incorrect.	$✓$
The probability of success is constant for each trial.	The probability of guessing the correct answer for each question is $\frac{1}{5} = 0.2,$ where $5$ is the number of options.	$✓$

Let

X

be a random variable representing the number of correct guesses out of the

5

trials. The possible values of

X

are

0,

1,

2,

3,

4,

and

5 .

Additionally, the probability of failure

q

will be

1 - 0.2 = 0.8 .

Substitute this information into the Binomial Probability Formula.

P (X = x) =_{n} C_{x} p^{x} q^{n - x} ⇓ P (X = x) =_{5} C_{x} (0.2)^{x} (0.8)^{5 - x}

The probability of guessing at least three questions correctly can be written as

P (X \geq 3) .

According to the Addition Rule of Probability, the sum of

P (X = 3),

P (X = 4),

and

P (X = 5)

is equal to the probability of

P (X \geq 3),

as there are exactly

5

questions. Begin by calculating

P (X = 3) .

P (X = x) =_{5} C_{x} (0.2)^{x} (0.8)^{5 - x}

Substitute

$x = 3$

P (X = 3) =_{5} C_{3} (0.2)^{3} (0.8)^{5 - 3}

Simplify right-hand side

$C (n, k) = \frac{n !}{k ! ( n - k ) !}$

P (X = 3) = (\frac{5 !}{3 ! ( 5 - 3 ) !}) (0.2)^{3} (0.8)^{5 - 3}

SubTerms

Subtract terms

P (X = 3) = (\frac{5 !}{3 ! \cdot 2 !}) (0.2)^{3} (0.8)^{2}

Write as a product

P (X = 3) = (\frac{5 \cdot 4 \cdot 3 !}{3 ! \cdot 2 \cdot 1 !}) (0.2)^{3} (0.8)^{2}

CancelCommonFac

Cancel out common factors

P (X = 3) = (\frac{5 \cdot 4 \cdot 3 !}{3 ! \cdot 2 \cdot 1 !}) (0.2)^{3} (0.8)^{2}

SimpQuot

Simplify quotient

P (X = 3) = (\frac{5 \cdot 4}{2 \cdot 1 !}) (0.2)^{3} (0.8)^{2}

Evaluate right-hand side

$1! = 1$

P (X = 3) = (\frac{5 \cdot 4}{2 \cdot 1}) (0.2)^{3} (0.8)^{2}

Multiply

P (X = 3) = (\frac{2 0}{2}) (0.2)^{3} (0.8)^{2}

CalcQuot

Calculate quotient

P (X = 3) = (10) (0.2)^{3} (0.8)^{2}

CalcPowProd

Calculate power and product

P (X = 3) = 0.0512

Follow a similar process to find

P (X = 4)

and

P (X = 5) .

$x$	$_{5} C_{x} (0.2)^{x} (0.8)^{5 - x}$	$P (X = x) =_{5} C_{x} (0.2)^{x} (0.8)^{5 - x}$
$3$	$_{5} C_{3} (0.2)^{3} (0.8)^{5 - 3} = \frac{1 6 0}{3 1 2 5}$	$P (X = 3) = 0.0512$
$4$	$_{5} C_{4} (0.2)^{4} (0.8)^{5 - 4} = \frac{2 0}{3 1 2 5}$	$P (X = 4) = 0.0064$
$5$	$_{5} C_{5} (0.2)^{5} (0.8)^{5 - 5} = \frac{1}{3 1 2 5}$	$P (X = 5) \approx 0.00032$

Now, add the probabilities to get

P (X \geq 3) .

P (X \geq 3) = P (X = 3) + P (X = 4) + P (X = 5)

SubstituteValues

Substitute values

P (X \geq 3) \approx 0.0512 + 0.0064 + 0.00032

AddTerms

Add terms

P (X \geq 3) \approx 0.05792

WritePercent

Convert to percent

P (X \geq 3) \approx 5.79 %

Therefore, the probability of Dylan guessing at least

3

correct answers is about

5.79 % .

Example

Finding the Number of Customers That Will Purchase Something

Izabella and Dylan are doing a fantastic job helping out at the clothing store. They noticed that the behavior of the customers can be modeled by a binomial experiment because there are two possible outcomes when customers enter the store — they either make a purchase or they do not.

Solve the following questions to help Izabella and Dylan find out if they can use their knowledge about binomial distributions in the store.

a Consider the following probability distributions.

Based on past experience, Magdalena knows that the probability that a customer will make a purchase is about

0.35 .

Suppose that

10

customers will enter the store in the next hour. Which graph describes the probability associated with the number of customers who will make a purchase?

b Suppose the probability that one customer will make a purchase changes to

0.5 .

What is the probability that exactly

3

of the next

8

customers will make a purchase? Round to two decimal places.

c Suppose that the store forecasts that

1500

customers will enter the store next month. If the probability that a customer will make a purchase is

0.35,

find the expected number of customers that make a purchase next month.

Hint

a Begin by identifying the conditions of a binomial experiment.

b Evaluate the Binomial Probability Formula at

x = 3 .

c The expected value is given by the product of the number of trials and the probability of success.

Solution

a Notice that the graphs each consider ten customers, so there are ten trials. In each trial, a success is that a customer makes a purchase. Start by identifying whether this experiment follows the three conditions of a binomial experiment.

Condition	Given Experiment	Is Satisfied?
There is a fixed number of independent trials.	There are $10$ trials — the $10$ customers that will enter the store.	$✓$
Each trial has two possible outcomes.	Whether a customer makes a purchase or not.	$✓$
The probability of success is constant for each trial.	There is a $0.35$ probability that a customer will make a purchase.	$✓$

Let

X

be the random variable representing the number of customers that will make a purchase. The possible values of

X

are

x = 0,

1,

\dots,

10 .

Note that the probability of failure

q

will be

1 - 0.35 = 0.65 .

This information can now be input into the Binomial Probability Formula.

P (X = x) =_{n} C_{x} p^{x} q^{n - x} ⇓ P (X = x) =_{10} C_{x} (0.35)^{x} (0.65)^{10 - x}

The binomial probability distribution will be determined by evaluating this formula for each of the possible values of

x .

In this case, the probability that none of the customers will make a purchase will be found first.

P (X = x) =_{10} C_{x} (0.35)^{x} (0.65)^{10 - x}

Substitute

$x = 0$

P (X = 0) =_{10} C_{0} (0.35)^{0} (0.65)^{10 - 0}

Evaluate right-hand side

$C (n, k) = \frac{n !}{k ! ( n - k ) !}$

P (X = 0) = (\frac{1 0 !}{0 ! ( 1 0 - 0 ) !}) (0.35)^{0} (0.65)^{10 - 0}

SubTerms

Subtract terms

P (X = 0) = (\frac{1 0 !}{0 ! ( 1 0 ! )}) (0.35)^{0} (0.65)^{10}

$0! = 1$

P (X = 0) = (\frac{1 0 !}{1 ( 1 0 ! )}) (0.35)^{0} (0.65)^{10}

MultByOne

$a \cdot 1 = a$

P (X = 0) = (\frac{1 0 !}{1 0 !}) (0.35)^{0} (0.65)^{10}

QuotOne

$\frac{a}{a} = 1$

P (X = 0) = 1 (0.35)^{0} (0.65)^{10}

CalcPowProd

Calculate power and product

P (X = 0) = 0.013462 \dots

RoundDec

Round to $4$ decimal place(s)

P (X = 0) \approx 0.0135

Similarly, the probability for the remaining values of

x

will be found.

$x$	$_{10} C_{x} (0.35)^{x} (0.65)^{10 - x}$	$P (X = x) =_{10} C_{x} (0.35)^{x} (0.65)^{10 - x}$
$0$	$_{10} C_{0} (0.35)^{0} (0.65)^{10 - 0}$	$P (X = 0) \approx 0.01346$
$1$	$_{10} C_{1} (0.35)^{1} (0.65)^{10 - 1}$	$P (X = 1) \approx 0.07249$
$2$	$_{10} C_{2} (0.35)^{2} (0.65)^{10 - 2}$	$P (X = 2) \approx 0.17565$
$3$	$_{10} C_{3} (0.35)^{3} (0.65)^{10 - 3}$	$P (X = 3) \approx 0.25222$
$4$	$_{10} C_{4} (0.35)^{4} (0.65)^{10 - 4}$	$P (X = 4) \approx 0.23767$
$5$	$_{10} C_{5} (0.35)^{5} (0.65)^{10 - 5}$	$P (X = 5) \approx 0.15357$
$6$	$_{10} C_{6} (0.35)^{6} (0.65)^{10 - 6}$	$P (X = 6) \approx 0.06891$
$7$	$_{10} C_{7} (0.35)^{7} (0.65)^{10 - 7}$	$P (X = 7) \approx 0.02120$
$8$	$_{10} C_{8} (0.35)^{8} (0.65)^{10 - 8}$	$P (X = 8) \approx 0.00428$
$9$	$_{10} C_{9} (0.35)^{9} (0.65)^{10 - 9}$	$P (X = 9) \approx 0.00051$
$10$	$_{10} C_{10} (0.35)^{10} (0.65)^{10 - 10}$	$P (X = 10) \approx 0.00003$

The table can now be used to graph the probability distribution.

The graphed distribution corresponds to option C.

b In order to find the probability that exactly

3

out of

8

customers will make a purchase,

P (X = 3)

needs to be found. In this case, the number of trials is

8,

the probability of success is

0.5,

and the probability of failure is

0.5 .

Use the Binomial Probability Formula again.

P (X = x) =_{8} C_{x} (0.5)^{x} (0.5)^{8 - x}

Substitute

$x = 3$

P (X = 3) =_{8} C_{3} (0.5)^{3} (0.5)^{8 - 3}

Evaluate right-hand side

$C (n, k) = \frac{n !}{k ! ( n - k ) !}$

P (X = 3) = (\frac{8 !}{3 ! \cdot ( 8 - 3 ) !}) (0.5)^{3} (0.5)^{8 - 3}

SubTerms

Subtract terms

P (X = 3) = (\frac{8 !}{3 ! \cdot 5 !}) (0.5)^{3} (0.5)^{5}

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

P (X = 3) = (\frac{8 !}{3 ! \cdot 5 !}) (0.5)^{8}

SplitIntoFactors

Split into factors

P (X = 3) = (\frac{8 \cdot 7 \cdot 6 \cdot 5 !}{3 \cdot 2 \cdot 1 \cdot 5 !}) (0.5)^{8}

CancelCommonFac

Cancel out common factors

P (X = 3) = (\frac{8 \cdot 7 \cdot 6 \cdot 5 !}{3 \cdot 2 \cdot 1 \cdot 5 !}) (0.5)^{8}

SimpQuot

Simplify quotient

P (X = 3) = (\frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}) (0.5)^{8}

Multiply

P (X = 3) = (\frac{3 3 6}{6}) (0.5)^{8}

CalcQuot

Calculate quotient

P (X = 3) = 56 (0.5)^{8}

CalcPowProd

Calculate power and product

P (X = 3) = 0.21875

RoundDec

Round to $2$ decimal place(s)

P (X = 3) \approx 0.22

There is a probability of about

0.22

that exactly

3

out of

8

customers will make a purchase.

c Magdalena estimates that about

1500

customers will enter the store next month. Let

X

be the random variable representing the number of customers that will make a purchase. Since this random variable follows a binomial distribution, its expected value is given by the product of the number of trials

n

and the probability of success

p .

\underline{Expected Value of X} E (X) = n p

In this case, the number of trials

n

1500

and the probability of success is

0.35 .

The expected number of customers who will make a purchase next month can be determined with this information.

E (X) = 1500 \cdot 0.35 ⇕ E (X) = 525

Therefore,

525

customers are expected to make a purchase next month.

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Catch-Up and Review

Example

Variance vs Standard Deviation

Hint

Solution

Collection I

Collection II

Conclusion

Hint

Solution

The Scratch-off Cards

Heights of 100 People

Rolling a Die

Blood Type O

Conclusion

Proof

The Probability for a Fixed Sequence

Finding the Total Number of Possible Sequences

Getting the Binomial Probability

Hint

Solution

Hint

Solution

Heights of $100$ People