Suppose that on average 2.5% of visitors to your website sign up for your newsletter. In a recent week, 2701 visitors out of a total of 108879 signed up.
Using a binomial distribution. What is the probability that number of visitors who signed up is 2701 or fewer?
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1 point

A.  0.125

B.  0.346

C.  0.414

D.  0.007

E.  0.015

To solve this problem using a binomial distribution, we need to apply the binomial cumulative probability formula to calculate the probability that 2701 or fewer visitors signed up for the newsletter.

### Step-by-step Solution:

1. **Parameters of the binomial distribution:**
   - \( n = 108879 \) (total number of visitors)
   - \( p = 0.025 \) (the probability of success, i.e., a visitor signing up)
   - \( k = 2701 \) (the number of successes, i.e., visitors who signed up)

The binomial distribution is given by:

\[
P(X \leq k) = \sum_{x=0}^{k} \binom{n}{x} p^x (1-p)^{n-x}
\]

However, for large values of \(n\), the binomial distribution can be approximated by a normal distribution because the binomial distribution tends to a normal distribution as \(n\) becomes large.

2. **Normal approximation:**
   - The **mean** (\(\mu\)) of the binomial distribution is:
     \[
     \mu = n \times p = 108879 \times 0.025 = 2721.975
     \]
   - The **standard deviation** (\(\sigma\)) of the binomial distribution is:
     \[
     \sigma = \sqrt{n \times p \times (1-p)} = \sqrt{108879 \times 0.025 \times (1 - 0.025)} \approx 51.415
     \]

3. **Convert to a standard normal variable:**
   Using the normal approximation, we can find the probability that \( X \leq 2701 \) by calculating the z-score:

   \[
   Z = \frac{X - \mu}{\sigma} = \frac{2701 - 2721.975}{51.415} \approx \frac{-20.975}{51.415} \approx -0.408
   \]

4. **Find the cumulative probability:**
   Using standard normal distribution tables or a calculator, the cumulative probability for \( Z = -0.408 \) is approximately 0.342.

Therefore, the probability that 2701 or fewer visitors signed up is about **0.342**.

### Closest answer: 
**B. 0.346**

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Would you like further details or clarifications?

### Related Questions:
1. How does the binomial distribution approach the normal distribution for large \(n\)?
2. How would the results change if the sign-up rate increased to 3%?
3. What is the probability that **exactly** 2701 visitors signed up using the binomial distribution?
4. How would a smaller visitor count affect the accuracy of the normal approximation?
5. Can the Poisson distribution be used here? Why or why not?

**Tip:** The normal approximation to the binomial distribution works best when both \(np\) and \(n(1-p)\) are greater than 5.

Suppose that on average 2.5% of visitors to your website sign up for your newsletter. In a recent week, 2701 visitors out of a total of 108879 signed up. Using a binomial distribution, what is the probability that the number of visitors who signed up is 2701 or fewer? A) 0.125 B) 0.346 C) 0.414 D) 0.007 E) 0.015

Calculate the probability that 2701 or fewer visitors signed up for a newsletter using a binomial distribution and normal approximation. In this problem, 2.5% of visitors typically sign up, and we analyze data from a sample of 108879 visitors. Suitable for undergraduate statistics students.

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