mars, inc. claims that 20% of its m&m plain candies are orange. a sample of 100 such candies is randomly…

mars, inc. claims that 20% of its m&m plain candies are orange. a sample of 100 such candies is randomly selected. find the mean and standard deviation for the number of orange candies in such groups of 100.\n\na. $mu = 20, sigma = 0.20$\nb. $mu = 0.020, sigma = 0.20$\nc. $mu = 20, sigma = 4.0$\nd. $mu = 0.20, sigma = 4.0$

mars, inc. claims that 20% of its m&m plain candies are orange. a sample of 100 such candies is randomly selected. find the mean and standard deviation for the number of orange candies in such groups of 100.\n\na. $mu = 20, sigma = 0.20$\nb. $mu = 0.020, sigma = 0.20$\nc. $mu = 20, sigma = 4.0$\nd. $mu = 0.20, sigma = 4.0$

Answer

Explanation:

Step1: Identify the distribution

This is a binomial - distribution problem, where $n = 100$ (number of trials/sample size) and $p=0.2$ (probability of success).

Step2: Calculate the mean

The mean of a binomial distribution is given by $\mu = np$. Substitute $n = 100$ and $p = 0.2$ into the formula: $\mu=100\times0.2 = 20$.

Step3: Calculate the standard deviation

The standard deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$. Substitute $n = 100$ and $p = 0.2$ into the formula: $\sigma=\sqrt{100\times0.2\times(1 - 0.2)}=\sqrt{100\times0.2\times0.8}=\sqrt{16}=4.0$.

Answer:

C. $\mu = 20,\sigma = 4.0$