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Section 3.5 Normal Approximation (D5)

In this section, we use the normal distribution to approximate the binomial probability distribution for large \(n\text{.}\)

Exploration 3.5.1. Flip Many Coins.

(a)

Run the following code to simulate 1000 events where you flip coinflips=10 coins, and record the number of resulting heads:

(b)

Repeat the above task but chance coinflips to 20, 30, 50, 100, 200. As you increase coinflips, what do you notice about the shape of the histogram?

Remark 3.5.1.

Given a binomial random variable \(X\) wnere \(np, n(1-p)\geq 10\text{,}\) that is, there are at least 10 expected successes and failures, \(X\) can be approximated by a normal random variable with parameters

\begin{equation*} \mu=np, \sigma=\sqrt{np(1-p)}. \end{equation*}

Activity 3.5.2. Flip Many Coins - Normal Approximation.

Suppose we flip 100 coins and let \(X\) denote the number of heads, as in Exploration 3.5.1. Let \(N_X\) denote the normal random variable which approximates \(X\) as in Remark 3.4.5.

(a)

Use Remark 3.4.5 to find the mean and standard deviation of the normal random variable \(N_X\) that approximates \(X\text{.}\)

(b)

Fix and run the following code to simulate 1000 events where you flip coinflips=1000 coins, and record the number of resulting heads, as well as display a normal curve over the histogram:

(c)

Fix and run the following code to see the probability that you get less than 45 heads:

(d)

Fix and run the following code to find a value \(k\) so that the probability of getting more than \(k\) heads is 25%:

(e)

Do your outcomes make sense when looking at the histogram and curve?

Activity 3.5.3. Survey Responses.

A research group emails 1000 city employees with a survey, each recipient has a 20% chance of responding to the survey. Let \(X\) denote the number of employees who respond. Use the normal approximation Remark 3.4.5 to answer the following.

(a)

What is the mean and standard deviation of the normal random variable that approximates \(X\text{?}\)

(d)

Use Remark 3.4.5 to determine a value \(k\) so that there is a 95% chance between \(200-k\) and \(200+k\) employees respond to the survey.

Hint. Desmos

(e)

Run the following code to simulate 10000 events where you send surveys=1000 surveys, and record the number of resulting responses, as well as display a normal curve over the histogram:

Do your results make sense when looking at the histogram and curve?