To solve the problem, we need to calculate the probability that 6 or more customers out of 15 used their phones for guidance on purchasing decisions. Since this follows a binomial distribution, the formula for the probability of exactly $k$ successes in $n$ trials is:

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

Where:

• $n = 15$ (the number of customers),
• $p = 0.58$ (the probability that a customer uses their phone for guidance),
• $k$ is the number of successes.

We're interested in the cumulative probability for $X \geq 6$. We calculate this as:

$P(X \geq 6) = 1 - P(X \leq 5)$

Let's start by calculating $P(X \leq 5)$, which is the sum of probabilities from $k = 0$ to $k = 5$.

$P(X \leq 5) = \sum_{k=0}^{5} \binom{15}{k} \cdot 0.58^k \cdot (1 - 0.58)^{15-k}$

Once we have $P(X \leq 5)$, we subtract it from 1 to get $P(X \geq 6)$.

Let me calculate this.The probability that 6 or more customers out of 15 used their phones for guidance on purchasing decisions is approximately 0.9521 or 95.21%.

Would you like to go into more detail on any part of this calculation, or do you have any further questions?

Here are some related questions you might find interesting:

1. What is the probability that exactly 8 customers used their phones for guidance?
2. How does the probability change if the sample size increases to 20?
3. What is the probability that fewer than 10 customers used their phones for guidance?
4. How would the result change if the probability of using phones increased to 0.7?
5. What are the odds of exactly half of the customers using their phones?

Tip: In a binomial distribution, probabilities are calculated by summing up individual probabilities for a range of outcomes, making cumulative distribution functions (CDFs) very useful for determining "at least" or "at most" probabilities.