according to data from the american medical association, 10% of us are left handed. If three people, are randomly selected find the probability that they are all left handed. If groups of 50 people are randomly selected, what is the mean and stadard deviation of left handed people.
For the first part:
You can use a binomial probability table, or calculate by hand using the following formula: P(x) = (nCx)(p^x)[q^(n-x)]
For the second part:
Use the binomial approximation to a normal distribution.
mean = np = 50 * .10 = ?
standard deviation = √npq = (√(50 * .10 * .90) = ?
Note: q = 1 - p
Finish the calculation.
I hope these few hints will help.
To find the probability that all three people selected are left-handed, we need to calculate the probability of selecting one left-handed person three times in a row.
Given that 10% of the population is left-handed, the probability of selecting a left-handed person is 0.10.
Since each selection is independent of the others, the probability of selecting three left-handed people in a row is calculated by multiplying the individual probabilities together:
Probability of selecting a left-handed person = 0.10
Probability of selecting three left-handed people in a row = (0.10) * (0.10) * (0.10) = 0.001
Therefore, the probability that all three randomly selected people are left-handed is 0.001 or 0.1%.
Now, to calculate the mean and standard deviation of left-handed people in groups of 50:
Given that 10% of the population is left-handed, in a group of 50 people, the expected number of left-handed individuals would be:
Mean = 0.10 * 50 = 5
The variance is calculated by multiplying the probability of being left-handed (0.10) with the probability of not being left-handed (1 - 0.10 = 0.90) and then multiplying by the sample size:
Variance = (0.10) * (0.90) * 50 = 4.5
Thus, the standard deviation is the square root of the variance:
Standard deviation = √4.5 ≈ 2.12
Therefore, the mean number of left-handed people in groups of 50 is 5, and the standard deviation is approximately 2.12.
To calculate the probability that all three people randomly selected are left-handed, we can use the concept of independent events. Since the selection of each person is independent, we can multiply the probabilities together.
Given that 10% of the population is left-handed, the probability of selecting a left-handed person is 0.10, or 10%.
P(all three are left-handed) = P(left-handed) * P(left-handed) * P(left-handed)
= 0.10 * 0.10 * 0.10
= 0.001 or 0.1%
Therefore, the probability that all three randomly selected people are left-handed is 0.1%.
To calculate the mean and standard deviation of left-handed people in groups of 50, we first need to determine the expected number of left-handed individuals.
Given that the overall proportion of left-handed people is 10%, the expected number of left-handed individuals in a group of 50 can be found by multiplying the proportion by the group size.
Mean = Proportion * Group Size
= 0.10 * 50
= 5
Therefore, the mean number of left-handed individuals in a group of 50 is 5.
To calculate the standard deviation, we can use the formula for the standard deviation of a binomial distribution:
Standard Deviation = sqrt(n * p * (1 - p))
where n is the group size and p is the proportion of left-handed individuals.
Standard Deviation = sqrt(50 * 0.10 * (1 - 0.10))
= sqrt(50 * 0.10 * 0.90)
≈ sqrt(4.5)
≈ 2.12 (rounded to two decimal places)
Therefore, the standard deviation of left-handed individuals in a group of 50 is approximately 2.12.