Binomial distribution is a discrete probability distribution of the number of successes in a sequence of \(n\) independent experiments, each with its own two outcomes: success (with probability \(p\)) failure (with probability \(q=1-p\)).

Binomial distribution is determined by parameters \(n\) and \(p\).

• Bernoulli Trials
• Combination coefficient formula – \(n\) choose \(k\)
• Binomial distribution formula (pdf)
• Cumulative density distribution (cdf)
• Mean, Variance, and Standard deviation of Binomial Distribution
• Unusual Observations
• Histogram of Binomial Distribution (skewness, bell-shaped and \(𝑛\), \(p\) )

- Binary Data — For each trial or observation, there are two possible outcomes (dichotomous responses) for example, Success/Failure; Accept/Decline an offer; vote Yes/No

• Bernoulli Trials — For repeated \(𝑛\) trials

  • Each of \(𝑛\) trials has two possible outcomes. The outcome of interest is called a success (S), the other one as a failure (F)
  • Each trial has the same probability of getting \(S\). Let \(P(S)=p\), then $P(F)=q=1-p
  • The \(n\) trials are independent. The probability of success in a trial remains unchanged given the outcomes of all other trials.

• Binomial Random Variable — For Bernoulli trials of 𝑛 times, in which we observe a binary data, let \(𝑋\) be the number of getting success, then \(X\) is a binomial random variable, and the possible values \(x=0, 1, \dots,n\)

• Binomial Distribution — the probability distribution for Binomial Random Variable— there will be a formula for this

Assume that 35 % of people do not use iPad. Suppose we randomly select four (4) individuals, What is the probability that exactly one (1) of them does not use iPad?

Let’s call these people Allen (A), Brittany (B), Caroline (C), and Damian (D). Each one of the four scenarios below will satisfy the condition of “exactly 1 of them do not use iPad”:

• Scenario 1: A does not, B, C, D use iPad: 0.35×0.65×0.65×0.65=0.0961
• Scenario 2: B does not, A, C, D use iPad: 0.65×0.35×0.65×0.65=0.0961
• Scenario 3: C does not, A, B, D use iPad: 0.65×0.65×0.35×0.65=0.0961
• Scenario 4: D does not, A, B, C use iPad: 0.65×0.65×0.65×0.35=0.0961

The probability of exactly one 1 of 4 people does not use iPad is the sum of these probabilities. 0.0961 + 0.0961 + 0.0961 + 0.0961 = 4 × 0.0961 = 0.3844

The question from the prior slide asked for the probability of given number of successes, k, in a given number of trials, n, (k = 1 success in \(n=4\) trials), and we calculated this probability as

\[P(X=k) = {n \choose k}p^k(1-p)^{(n-k)} = \frac{n!}{k! \cdot (n-k)!}p^k(1-p)^{(n-k)}, k=0,1,2,...n\]

• In the probability formula for binomial distribution, \(𝑛!\) is read as \(𝒏\) factorial : \(𝑛!=1×2×3×…×𝑛\) for any positive inter $𝑛 $

• \(0!=1\)

• While \({n \choose k}\) is the combination, it shows the total possible ways of choosing \(𝑘\) items from a collection of \(𝑛\) items without ordering.

Example:

\({4\choose 1}=\frac {4!}{1!3!}=4\), say, take a quiz with 4 questions, there are four ways of getting exactly 1 correct: CIII, ICII, IICI, IIIC.

\({3\choose 2} =\frac{3!}{2!1!}=3\), say, given three fruits: an apple, an orange, and a pear, there are three ways that two fruits can be drawn from this set: apple and pear; apple and orange; orange and pear.

For any \(0≤𝑘≤𝑛\),

\[{n \choose k}= \frac{𝑛!}{𝑘!(𝑛−𝑘)!} \ \text{is equal to} \ {n \choose n-k} = \frac{𝑛!}{𝑘!(𝑛−𝑘)!}\]

In other words, \(𝑛\) choose \(𝑘\) is the same as not choose \(𝑛−𝑘\).

Example

1. When \(𝑘=0, {n \choose 0} = {n \choose n} = \frac{𝑛!}{𝑛!0!}=1\)

2. When \(𝑘=1, {n \choose 1} = {n \choose n-1} = \frac{𝑛!}{𝑛-1!1!}=n\)

3. Then, \({7\choose 1}={7\choose 6}, {7\choose 2} = {7\choose 5}, \ \text{and}\ {7\choose 3} = {7\choose 4}\)

Example: Historical data shows that about 51% of newborn babies are boys. If we randomly select four newborns, build the binomial model for the possible number of boys.

choose(4,0) * 0.51^0 * (1-0.51)^4 #calculates P(X=0)
choose(4,1) * 0.51^1 * (1-0.51)^3 #calculates P(X=1)
P_x_le_1 = 0.0576 + 0.2400; P_x_le_1 #calculates P(X<=1)

Example: Historical data shows that about 51% of newborn babies are boys. If we randomly select four newborns, build the binomial model for the possible number of boys.

dbinom(x = 0, size = 4, prob = 0.51) #calculates P(X=0)
dbinom(1, 4, 0.51) #calculates P(X=1); we can drop the argument label
pbinom(1, 4, 0.51, lower.tail = TRUE) #calculates P(X<=1)

Which of the following is not a condition that needs to be met for the binomial distribution to be applicable?

1. The trials must be independent.
2. The number of trials, 𝑛, must be fixed.
3. Each trial outcome must be classified as a success or a failure.
4. The number of desired successes, 𝑘, must be greater than the number of trials.
5. The probability of success, 𝑝, must be the same for each trial.

Which of the following is not a condition that needs to be met for the binomial distribution to be applicable?

1. The trials must be independent.
2. The number of trials, 𝑛, must be fixed.
3. Each trial outcome must be classified as a success or a failure.
4. The number of desired successes, 𝑘, must be greater than the number of trials.
5. The probability of success, 𝑝, must be the same for each trial.

Assume that there’s an insurance agency where 70% of individuals do not exceed their deductible. What is the probability that 3 of 8 randomly selected individuals will have exceeded the deductible, i.e. that 5 of 8 will not exceed the deductible?

Thus, the probability that 3 of 8 randomly selected individuals will have exceeded the insurance deductible is 0.2541 or 25.41%.

Assume that there’s an insurance agency where 70% of individuals do not exceed their deductible. What is the probability that 3 of 8 randomly selected individuals will have exceeded the deductible, i.e. that 5 of 8 will not exceed the deductible?

dbinom(x = 5, size = 8, prob = 0.7) #calculates P(X=5)
dbinom(5, 8, 0.7) #we can drop the argument label

dbinom(3, 8, 0.3) #we can drop the argument label

For binomial distribution with 𝑛 and 𝑝, it can be proved that \[\text{Mean} = \mu = np\] \[\text{Variance} =\sigma^2 = np(1-p)\] \[\text{The standard deviation} = \sigma = \sqrt{np(1-p)}\]

But this doesn’t mean in every random sample of 100 people, exactly 26.2 will be obese. In fact, that’s not even possible. In some samples, this value will be less, and in others more.

How much would we expect this value to vary?

In previous slide, the mean was 26.2, and the standard deviation was 4.4, we can calculate a range for the plausible number of obese Americans in random samples of 100.

\[𝜇±2𝜎=26.2 ± (2 × 4.4) = (17.4, 35)\] So, for a random sample of 100 Americans, the percent of number of obese Americans are usually between 17.4 and 35.

The percent either below 17.4 or higher than 35 would be considered unusual.

We may use a web applet to explore. Go to https://istats.shinyapps.io/BinomialDist/ and choose your “the number of Bernoulli Trials(n)” and “probability of Success (p)” or you may enter 𝑛 and 𝑝.

• Try different 𝑛, with 𝑝=0.5

• Try p close to 0 (𝑝=0.1) or 1(𝑝=0.95) with different 𝑛 (small or large)

Further considerations: - What happens to the shape of the distribution as 𝑛 stays constant and 𝑝 changes?

• What happens to shape of the distribution as 𝑝 stays constant and 𝑛 changes?

1. 𝑛 = 100, 𝑝 = 0.95 → 100 × 0.95 = 95; 100 × 0.05 = 5
2. 𝑛 = 25, 𝑝 = 0.45 → 25 × 0.45 = 11.25; 25 × 0.55 = 13.75
3. 𝑛 = 150, 𝑝 = 0.05 → 150 × 0.05 = 7.5; 150 × 0.95 = 142.5
4. 𝑛 = 500, 𝑝 = 0.015 → 500 × 0.015 = 7.5; 500× 0.985 = 492.5