Let's break down the problem step by step.

Given:

• Probability of an incoming call involving a fax: $p = 0.40$
• Sample size: $n = 15$
• Number of calls is modeled using a Binomial Distribution: $X \sim \text{Binomial}(n = 15, p = 0.40)$

Part (a) - Expected Value

The expected number of calls involving a fax is given by the formula for the expected value of a binomial distribution: $E(X) = n \cdot p$ Substitute the values of $n$ and $p$: $E(X) = 15 \cdot 0.40 = 6$

So, the expected number of fax-related calls is: $E(X) = 6$

Part (b) - Standard Deviation

The standard deviation for a binomial distribution is given by: $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$ Substitute the values: $\sigma = \sqrt{15 \cdot 0.40 \cdot (1 - 0.40)} = \sqrt{15 \cdot 0.40 \cdot 0.60}$ Let's calculate that.

$\sigma = \sqrt{15 \cdot 0.40 \cdot 0.60} = \sqrt{3.6} \approx 1.897$

So, the standard deviation is approximately: $\sigma \approx 1.897$

Part (c) - Probability of Exceeding the Expected Number by More than 2 Standard Deviations

We need to find the probability that the number of calls exceeds the expected value by more than 2 standard deviations. This means we are interested in:

$P(X > E(X) + 2\sigma)$

Using the values from parts (a) and (b): $E(X) + 2\sigma = 6 + 2 \times 1.897 \approx 6 + 3.794 = 9.794$

So we want to calculate $P(X > 9.794)$, which is the probability that more than 9.794 calls involve a fax. Since $X$ is discrete, this is equivalent to finding $P(X \geq 10)$.

We will calculate $P(X \geq 10)$ using the binomial distribution formula or a binomial calculator.

Let me calculate that for you.

The probability that the number of calls involving a fax exceeds the expected number by more than 2 standard deviations is approximately:

$P(X \geq 10) \approx 0.034$

Summary of Results:

• (a) Expected number of fax-related calls: $E(X) = 6$
• (b) Standard deviation: $\sigma \approx 1.897$
• (c) Probability that the number of calls exceeds the expected number by more than 2 standard deviations: $P(X \geq 10) \approx 0.034$

Do you want more details or have any other questions?

Here are 5 related questions to explore:

1. How would the expected number change if the probability of a fax call was 60%?
2. What is the probability that exactly 8 calls involve a fax transmission?
3. What is the cumulative probability that at most 5 calls involve a fax transmission?
4. How would the standard deviation change for a sample of 25 calls?
5. How can we approximate the binomial distribution using a normal distribution for large $n$?

Tip: The binomial distribution is symmetric when $p = 0.5$, but becomes skewed when $p$ is far from 0.5.