A candy company claims that 11% of its plain candies are orange, and a sample of 200 such candies is randomly selected
a. find the mean and standard deviation for the number of orange candies in such groups of 100. use the formulas below to calculate
the mean and standard deviation of the distribution. μ = np σ = √npq
To find the mean and standard deviation for the number of orange candies in groups of 100, you can use the formulas for mean and standard deviation of a binomial distribution.
Mean (μ) = np
Standard Deviation (σ) = √(npq)
In this case, p represents the probability of selecting an orange candy, q represents the probability of selecting a candy that is not orange, and n represents the sample size.
Given that the candy company claims that 11% of its plain candies are orange, we can determine that p = 0.11. Since p is the probability of success, q can be calculated as q = 1 - p = 1 - 0.11 = 0.89.
Now we can calculate the mean and standard deviation using the formulas.
Mean (μ) = np = 100 * 0.11 = 11
Standard Deviation (σ) = √(npq) = √(100 * 0.11 * 0.89) ≈ √9.779 ≈ 3.12
Therefore, the mean number of orange candies in groups of 100 is 11, and the standard deviation is approximately 3.12.