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

0.784.
​(Round to three decimal places as​ needed.)
Part 3
​c) What is the probability that there are no more than 10 ​righties?
The probability that there are no more than 10 righties is

0.431.
​(Round to three decimal places as​ needed.)
Part 4
​d) What is the probability that there are exactly 6 of​ each?
The probability that there are exactly 6 of each is

enter your response here
​(Do not round until the final answer. Then round to five decimal places as​ needed.)

To solve part **d** (the probability that there are exactly 6 right-handed and 6 left-handed people in the sample), we can model this as a binomial distribution problem, where the probability of success (being right-handed) is \( p = 0.88 \), the probability of failure (being left-handed) is \( 1 - p = 0.12 \), and the sample size is \( n = 12 \).

The probability of exactly \( k \) right-handed people in a sample of \( n \) is given by the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}
\]

For this problem, we want to find \( P(X = 6) \), where \( X \) is the number of right-handed people. This means:

\[
P(X = 6) = \binom{12}{6} (0.88)^6 (0.12)^6
\]

Now let's calculate this.The probability that there are exactly 6 right-handed and 6 left-handed people in the sample is approximately \( 0.00128 \) (rounded to five decimal places).

Would you like more details or have any further questions?

Here are five related questions to explore:
1. How would the probability change if the sample size increased to 20?
2. What is the probability of having exactly 8 right-handed people in the sample of 12?
3. Can we approximate this binomial distribution using a normal distribution?
4. How does the mean number of right-handers change if the sample size is reduced to 10?
5. What is the probability that more than 10 out of 12 people are right-handed?

**Tip:** The binomial distribution is ideal for modeling events with only two possible outcomes, such as success/failure or yes/no scenarios.

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.
a) Find the mean and standard deviation of the number of​ right-handers in the group.
b) What is the probability that they are not all​ right-handed?
c) What is the probability that there are no more than 10 ​righties?
d) What is the probability that there are exactly 6 of​ each?

Explore how to calculate the mean, standard deviation, and probabilities for a binomial distribution with a sample of 12 customers. Learn how to find the probability that not all customers are right-handed, the probability of having no more than 10 right-handed users, and the chance of having exactly 6 right-handed and 6 left-handed customers. Suitable for high school students studying probability and statistics.

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