To find the parameter $p$ of the Binomial distribution that matches the given probabilities, we'll use the properties of the Binomial distribution.

The Binomial distribution is defined as:

$P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$

Given:

• $n = 5$
• The probabilities $P(X = x)$ for $x = 0, 1, 2, 3, 4, 5$ are provided.

To find $p$, we should use the fact that for a Binomial distribution:

$\text{E}(X) = np$

We can calculate the expected value $\text{E}(X)$ using the provided distribution and then set it equal to $5p$ (since $n = 5$).

Step 1: Calculate the Expected Value $\text{E}(X)$

$\text{E}(X) = \sum_{x=0}^{5} x \cdot P(X = x)$

Substituting the given probabilities:

$\text{E}(X) = 0 \cdot 0.01 + 1 \cdot 0.09 + 2 \cdot 0.25 + 3 \cdot 0.34 + 4 \cdot 0.24 + 5 \cdot 0.07$

$\text{E}(X) = 0 + 0.09 + 0.5 + 1.02 + 0.96 + 0.35 = 2.92$

Step 2: Find $p$

Given $\text{E}(X) = np$:

$2.92 = 5p$

$p = \frac{2.92}{5} = 0.584$

Thus, the parameter $p$ for this Binomial distribution is approximately 0.584.

Would you like more details on this process or have any questions?

Here are 5 related questions for further exploration:

1. How would the distribution change if the parameter $p$ was slightly higher?
2. What is the variance of $X$ for this Binomial distribution?
3. How does the Binomial distribution relate to the Normal distribution for large $n$?
4. How would you calculate the probability of the equipment working for exactly 4 days?
5. What does the standard deviation tell you about the reliability of the equipment?

Tip: When dealing with distributions, understanding both the mean and the variance is crucial for interpreting the behavior of random variables.