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

$X = Sum of Two Dice$
$x$	Frequency	$P (X = x)$
$2$	$1$	$\frac{1}{3 6} \approx 0.028$
$3$	$2$	$\frac{2}{3 6} \approx 0.056$
$4$	$3$	$\frac{3}{3 6} \approx 0.083$
$5$	$4$	$\frac{4}{3 6} \approx 0.111$
$6$	$5$	$\frac{5}{3 6} \approx 0.139$
$7$	$6$	$\frac{6}{3 6} \approx 0.167$
$8$	$5$	$\frac{5}{3 6} \approx 0.139$
$9$	$4$	$\frac{4}{3 6} \approx 0.111$
$10$	$3$	$\frac{3}{3 6} \approx 0.083$
$11$	$2$	$\frac{2}{3 6} \approx 0.056$
$12$	$1$	$\frac{1}{3 6} \approx 0.028$

Number of Heads, $x$	Tally	Frequency
$0$	$∣ ∣ ∣ ∣ ∣$	$6$
$1$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$22$
$2$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$37$
$3$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$28$
$4$	$∣ ∣ ∣ ∣ ∣ ∣$	$7$

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$0.0625$	$0.25$	$0.375$	$0.25$	$0.0625$

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$0.06$	$0.22$	$0.37$	$0.28$	$0.07$

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
Possible Outcomes	$T T T T$	$T T T H,$ $T T H T,$ $T H T T,$ $H T T T$	$H H T T,$ $H T T H,$ $T T H H,$ $H T H T,$ $T H T H,$ $T H H T$	$H H H T,$ $H H T H,$ $H T H H,$ $T H H H$	$H H H H$
Frequency	$1$	$4$	$6$	$4$	$1$

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
Frequency	$1$	$4$	$6$	$4$	$1$
$P (X = x)$	$\frac{1}{1 6}$	$\frac{4}{1 6} = \frac{1}{4}$	$\frac{6}{1 6} = \frac{3}{8}$	$\frac{4}{1 6} = \frac{1}{4}$	$\frac{1}{1 6}$

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$0.0625$	$0.25$	$0.375$	$0.25$	$0.0625$

Number of Heads, $x$	Tally	Frequency
$0$	$∣ ∣ ∣ ∣ ∣$	$6$
$1$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$22$
$2$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$37$
$3$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$28$
$4$	$∣ ∣ ∣ ∣ ∣ ∣$	$7$

Number of Heads, $x$	Tally	Frequency	$P (X = x)$
$0$	$∣ ∣ ∣ ∣ ∣$	$6$	$\frac{6}{1 0 0}$
$1$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$22$	$\frac{2 2}{1 0 0}$
$2$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$37$	$\frac{3 7}{1 0 0}$
$3$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$28$	$\frac{2 8}{1 0 0}$
$4$	$∣ ∣ ∣ ∣ ∣ ∣$	$7$	$\frac{7}{1 0 0}$

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$0.06$	$0.22$	$0.37$	$0.28$	$0.07$

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

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

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

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

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	$\approx 1864.20$

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

$x$	$_{5} C_{x} (\frac{1}{4})^{x} (\frac{3}{4})^{5 - x}$	$P (X = x) =_{5} C_{x} (\frac{1}{4})^{x} (\frac{3}{4})^{5 - x}$
$0$	$_{5} C_{0} (\frac{1}{4})^{0} (\frac{3}{4})^{5 - 0} = \frac{2 4 3}{1 0 2 4}$	$\approx 0.237$
$1$	$_{5} C_{1} (\frac{1}{4})^{1} (\frac{3}{4})^{5 - 1} = \frac{4 0 5}{1 0 2 4}$	$\approx 0.396$
$2$	$_{5} C_{2} (\frac{1}{4})^{2} (\frac{3}{4})^{5 - 2} = \frac{2 7 0}{1 0 2 4}$	$\approx 0.264$
$3$	$_{5} C_{3} (\frac{1}{4})^{3} (\frac{3}{4})^{5 - 3} = \frac{9 0}{1 0 2 4}$	$\approx 0.088$
$4$	$_{5} C_{4} (\frac{1}{4})^{4} (\frac{3}{4})^{5 - 4} = \frac{1 5}{1 0 2 4}$	$\approx 0.015$
$5$	$_{5} C_{5} (\frac{1}{4})^{5} (\frac{3}{4})^{5 - 5} = \frac{1}{1 0 2 4}$	$\approx 0.001$

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.	$✓$

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

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.	$✓$

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

Read each situation carefully.

	Description
Situation I	Determining the number of girls in a family with $6$ children if every newborn baby has a probability of about $0.45$ of being male and $0.55$ of being female.
Situation II	Removing cards from a standard deck of cards and recording whether each one is red or black. No replacement occurs.
Situation III	Ali records the number of days that it rains in September if there is a $37 %$ of probability that it will rain each day in Salt Lake City.

Which of these situations can be modeled as a binomial experiment? Select all that apply.

We want to identify whether we can model the given situations with binomial experiments. Before we analyze each situation individually, let's review the conditions a binomial experiment must 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.

Let's begin by analyzing Situation I.

Situation I

In this situation, we want to determine the number of girls in a family with 6 children.

Because each child can be a girl or a boy, we have two possible outcomes. The success of this experiment will be a child being a girl. By assuming that the probability of a newborn baby being a girl is independent from any previous pregnancies, we can set the probability of success to be 0.55. P(Success)=0.55 Finally, because the family has 6 children, we have a fixed number of independent trials. This means that this situation can be modeled as a binomial experiment.

Situation II

This situation involves removing cards from a standard deck without replacing them and recording whether each card is red or black.

This situation might be able to be considered as a binomial experiment because there are only two possible outcomes — each card will be either black or red. Also, the number of trials is fixed because a standard deck has exactly 52 cards.

Number of Trials	Possible Outcomes
52	Red or Black

However, let's see if the probability of success in each trial is constant. There are 52 cards in the deck, half of which are red. Therefore, the probability of the first drawn card being red is as follows. P(first card is red)=26/52=1/2 After we remove the first card from the deck, there are 51 cards left. Suppose that the first removed card was, in fact, red and that are now 26−1=25 red cards remaining in the deck. We can use these values to calculate the probability of the second card being red. P(second card is red)=25/51 Since 2551≠ 12, the probability of success in the second trial is different from the probability of success in the first trial. Therefore, the third condition is not satisfied. We can conclude that this situation does not represent a binomial experiment.

Situation III

Finally, let's analyze whether recording the number of days that it rains during September in Salt Lake City can be reduced to a binomial experiment.

We can see that the possible outcomes are that it rains or it does not. If we consider a rainy day a success, its probability of rain each day is 37 %, or 0.37. Moreover, the number of trials is given by the number of days in September.

Number of Trials	Possible Outcomes
30	Rain and no rain

Finally, if we assume that the probability of rain each day is independent from any other day, we can say that this situation can be represented as a binomial experiment.

Conclusion

We have analyzed each situation individually. The table summarizes whether or not each situation can be represented by a binomial experiment.

	Binomial?
Situation I	Yes ✓
Situation II	No *
Situation III	Yes ✓

From historical data, a company knows there is a $45 %$ chance that a person visiting their website will make a purchase.

Webpage of the company opened in a laptop

What is the probability that

10

out of the next

20

visitors will make a purchase? Round to three decimal places.

We want to determine the probability that 10 out of the next 20 visitors to the company's website will make a purchase. We can see that this situation seems to be a binomial experiment because there are only two outcomes — a visitor will either make a purchase or will not. Let's see if this situation meets the other conditions of a binomial experiment.

Condition	Given Experiment	Is Satisfied?
There is a fixed number of independent trials.	There are 20 trials — the 20 visitors at the website.	✓
Each trial has two possible outcomes.	Whether a visitor makes a purchase or not.	✓
The probability of success is constant for each trial.	There is a 0.45 probability that a visitor will make a purchase.	✓

We have determined that this situation can be modeled as a binomial experiment. This means that we can use the Binomial Probability Formula to calculate the probability that exactly 10 out of the 20 customers will make a purchase. P(X=x)=_nC_xp^xq^(n-x) In this case, X is the random variable representing the number of customers that will make a purchase. The number of trials n is 20. The probability of success p, which is a visitor making a purchase, is 0.45, and the probability of failure q is 1− 0.45= 0.55, and x= 10. P(X=x)=_nC_xp^xq^(n-x) ⇓ P(X= 10)=_(20)C_(10)( 0.45)^(10)( 0.55)^(20- 10) Now we can evaluate the formula to determine the value of P(X= 10).

Therefore, the probability that 10 out of the next 20 visitors to the website make a purchase is about 0.159, or 15.9 %.

A bag contains $20$ red marbles and $25$ blue marbles.

Suppose that

6

marbles are taken out of the bag, one at a time. Each marble is replaced before the next is drawn. What is the probability of getting at most

3

blue marbles? Round to four decimal places. Write the answer in percent form.

We want to determine the probability of getting at most 3 blue marbles when drawing 6 marbles, with replacement, from the bag. Let X be the random variable representing this situation. X=Number of blue marbles There are two possible outcomes: the drawn marble is either red or blue. Because we replace every marble we draw before we take out the next one, the number of marbles of each color remains the same. As a result, the probability of choosing a red or a blue marble is constant in every trial. To calculate these probabilities, let's first calculate the total number of marbles. Total Number of Marbles= 25+ 20 ⇕ Total Number of Marbles= 45 This means that the probability of success p, drawing a blue marble, is 25 45, and the probability of failure q is 20 45. Since there are 6 trials in this experiment, we can use the Binomial Probability Formula to determine the probability of getting at most 3 blue marbles. P(X≤ 3)=? To find this probability, we can use the Addition Rule of Probability to find and add the probabilities of P(X=0), P(X=1), P(X=2), and P(X=3). Let's first find the probability of P(X=0).

We can follow a similar process to determine the probabilities of P(X=1), P(X=2), and P(X=3).

x	_6C_x(25/45)^x(20/45)^(6-x)	P(X=x)= _6C_x(25/45)^x(20/45)^(6-x)
0	_6C_0( 25/45)^0( 20/45)^(6- 0)	P(X= 0)≈ 0.0077
1	_6C_1( 25/45)^1( 20/45)^(6- 1)	P(X= 1)≈ 0.0578
2	_6C_2( 25/45)^2( 20/45)^(6- 2)	P(X= 2)≈ 0.1806
3	_6C_3( 25/45)^3( 20/45)^(6- 3)	P(X= 3)≈ 0.3011

We can add the values of the last column to get the probability of P(X≤3).

This means that the probability of getting at most three blue marbles is about 54.72 %.

The probability of flipping heads on a biased coin is $0.58 .$ Ignacio calculated the probability of getting four heads in six coin tosses of this biased coin.

Ignacio steps: Step 1. P(X=4)=6_C_4(0.58)^2(0.42)^4; Step 2. P(X=4)=6!/(4!2!)(0.58)^2(0.42)^4; Step 3. P(X=4)=15(0.58)^2(0.42)^4; Step 4. P(X=4)=0.16

However, there is a mistake in Ignacio's calculations. Which step contains the mistake?

To find the mistake in the given procedure, we will evaluate the Binomial Probability Formula step by step and compare it with the ones made by Ignacio. Consider the Binomial Probability Formula. P(X=x)=_nC_xp^xq^(n-x) In this case, X is the random variable representing the number of heads results in six coin tosses. We want to determine the probability of P(X= 4) given that there are 6 trials. Because the probability of success p, getting heads, is 0.58, the probability of failure is 0.42. Let's substitute this information into the formula.

Now let's compare our expression with the one Ignacio wrote in Step 1.

Our Expression	Ignacio's Expression
P(X=4)=_6C_4(0.58)^4*(0.42)^2	P(X=4)=_6C_4(0.58)^2*(0.42)^4

We can see that the exponents are switched. This means that Ignacio made a mistake in Step 1 when calculating the exponents. Now, let's calculate the probability using the correct expression.

Therefore, the probability Ignacio was looking for is about 0.30, or 30 %. Ignacio's calculations were wrong because of the mistake was made in Step 1 when he switched the exponents.

	13 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Probability and Binomial Distributions of Data

Catch-Up and Review

Example

Answer

Hint

Solution

Example

Variance vs Standard Deviation

Hint

Solution

Collection I

Collection II

Conclusion

Example

Extra

Hint

Solution

The Scratch-off Cards

Heights of 100 People

Rolling a Die

Blood Type O

Conclusion

Example

Extra

Proof

The Probability for a Fixed Sequence

Finding the Total Number of Possible Sequences

Getting the Binomial Probability

Hint

Solution

Hint

Solution

Situation I

Situation II

Situation III

Conclusion

Probability and Binomial Distributions of Data

Recommended exercises

Heights of $100$ People