We are working with a binomial probability distribution where:

• $p = 0.88$ (probability of being right-handed)
• $n = 12$ (the sample size, or number of customers)

Part a) Mean and standard deviation

The mean $\mu$ and standard deviation $\sigma$ for a binomial distribution can be calculated using the following formulas:

• Mean: $\mu = n \cdot p$
• Standard deviation: $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$

The values for part a are already provided:

• Mean $\mu = 10.56$
• Standard deviation $\sigma = 1.13$

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Part b) What is the probability that they are not all right-handed?

To find the probability that not all of the 12 customers are right-handed, we first calculate the probability that they are all right-handed. This occurs when all 12 customers are right-handed, which is a binomial probability where $X = 12$.

Using the formula for binomial probability:

$P(X = 12) = p^{12} = (0.88)^{12}$

Now, to find the probability that not all customers are right-handed, we use the complement rule:

$P(\text{Not all right-handed}) = 1 - P(X = 12)$

Let’s compute that.The probability that not all of the customers are right-handed is approximately $0.784$ (rounded to three decimal places).

Would you like further details or have any questions? Here are five related questions for further practice:

1. What is the probability that exactly 10 customers are right-handed?
2. What is the probability that fewer than 10 customers are right-handed?
3. What is the probability that more than 10 customers are right-handed?
4. How would the mean and standard deviation change if the sample size increased to 20 customers?
5. What is the probability that exactly 11 customers are right-handed?

Tip: In a binomial distribution, always check if you can use the complement rule for easier probability calculations.