We are working with a binomial distribution problem where:

• $n = 12$ (sample size),
• $p = 0.85$ (probability of a customer being right-handed),
• $q = 1 - p = 0.15$ (probability of a customer being left-handed).

Part (c): Finding the probability that there are no more than 10 right-handers

To calculate the probability that there are no more than 10 right-handers, we need to find:

$P(X \leq 10)$

Where $X$ is the number of right-handed customers in the sample, which follows a binomial distribution:

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

We can compute the cumulative probability for $P(X \leq 10)$ by summing the probabilities for all possible values of $X$ from 0 to 10:

$P(X \leq 10) = P(X = 0) + P(X = 1) + \dots + P(X = 10)$

Let me calculate that for you.The probability that there are no more than 10 right-handers in the group is approximately $0.557$ (rounded to three decimal places).

Would you like further details or explanations?

Here are 5 related questions:

1. What is the probability that there are exactly 10 right-handers?
2. How would the probability change if the sample size increased to 20?
3. Can you calculate the probability that there are fewer than 8 right-handed customers?
4. What is the probability of having exactly 12 right-handed people?
5. How does the standard deviation affect the interpretation of this problem?

Tip: The binomial distribution is useful when you have a fixed number of independent trials, each with two possible outcomes.