According to Masterfoods, the company that manufactures M&M’s,
12% of peanut M&M’s are brown,
15% are yellow,
12% are red,
23% are blue,
23% are orange, and
15% are green.
You randomly select six peanut M&M’s from an extra-large bag of the candies.
If you repeatedly select random samples of six peanut M&M’s, on average how many do you expect to be red?
With what standard deviation?
I only need formulas the rest of the questions I've answered I am just stuck on these two because I am not sure what formula I need.
To calculate the expected number of red M&M's, you need to find the probability of selecting a red M&M in each trial and multiply it by the number of trials.
In this case, the probability of selecting a red M&M is 12%. Hence, the expected number of red M&M's in each sample of 6 candies is (12/100) * 6 = 0.12 * 6 = 0.72
Therefore, on average, you can expect to find approximately 0.72 red M&M's in each random sample of 6 peanut M&M's.
To calculate the standard deviation, you can use the formula for the variance of a binomial distribution: variance = np(1-p).
In this case, the probability of selecting a red M&M (p) is 12/100, and the number of trials (n) is 6.
So, the variance (σ²) = 6 * (12/100) * (1 - 12/100) = 6 * (0.12) * (0.88) = 0.6336
To obtain the standard deviation (σ), you take the square root of the variance:
σ = √0.6336 ≈ 0.796
Therefore, the standard deviation of the number of red M&M's in each sample of 6 candies is approximately 0.796.