A manufacturer of game controllers is concerned that their controller may be difficult for​ left-handed users. They set out to find lefties to test. Suppose that about 88​% of the population is​ right-handed. They select a sample of twelve customers at random in their stores. Complete parts a through e below.
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Part 1
​a) Find the mean and standard deviation of the number of​ right-handers in the group.
The mean number of righties is 
  
10.56.
The standard deviation would be 
  
1.13. 
​(Type integers or decimals rounded to two decimal places as​ needed.)
Part 2
​b) What is the probability that they are not all​ right-handed?
The probability that they are not all righthanded is 
  
enter your response here.
​(Round to three decimal places as​ needed.)

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 \( \sigma = 1.13 \)

---

### 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.

A manufacturer of game controllers is concerned that their controller may be difficult for left-handed users. They set out to find lefties to test. Suppose that about 88​% of the population is​ right-handed. They select a sample of twelve customers at random in their stores. Complete parts a through e below.

Part 1
​a) Find the mean and standard deviation of the number of​ right-handers in the group. The mean number of righties is 10.56. The standard deviation would be 1.13.

Part 2
​b) What is the probability that they are not all​ right-handed?

Explore a binomial distribution problem involving right-handed and left-handed customers in a group of 12. Learn how to calculate the mean, standard deviation, and the probability of not all customers being right-handed. Key concepts include the binomial probability formula and the complement rule.

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