91影视

Given in the question that, According to the Mars candy company, $20\%$of its plain M&M鈥檚 candies are orange. Assume that the company鈥檚 claim is true. Suppose that you reach into a large bag of plain M&M鈥檚 (without looking) and pull out $8$candies.

Probability of success $(p)$localid="1649747575227" $=20\%=0.20$.

Number of trials localid="1649747578779" $\left(n\right)=8$

The random variable $X$, which is a binomial random variable, is as follows:

• There are two types of successes and failures: orange candy and non-orange candy.
• The candies are kept apart from one another.
• The number of candies is predetermined.
• The likelihood of receiving orange candy is set.

Because all of the requirements for a binomial distribution are predetermined. As a result, the variable $X$is a random variable with a binomial distribution.

Given in the question that, According to the Mars candy company, $20\%$of its plain M&M鈥檚 candies are orange. Assume that the company鈥檚 claim is true. Suppose that you reach into a large bag of plain M&M鈥檚 (without looking) and pull out $8$
candies. Let $X$= the number of orange candies you get. We need to find and interpret the expected value of $X$.

The mean can be calculated as:

$渭=n脳p$

$=8(0.20)$

$=1.60$

The above computed mean shows that the expected number of candies is 1 .

Given in the question that, According to the Mars candy company, $20\%$of its plain M&M鈥檚 candies are orange. Assume that the company鈥檚 claim is true. Suppose that you reach into a large bag of plain M&M鈥檚 (without looking) and pull out $8$candies. Let $X$= the number of orange candies you get. We need to find and interpret the standard deviation of $X$.

The standard deviation can be calculated as:

$蟽=\sqrt{n脳p脳(1鈭�p)}$

$=\sqrt{8\left(0.20\right)(1鈭�0.20)}$

$=1.1314$

The above-computed mean indicates that the number of candies will vary by $1.1314$by the average of $1.60$candies.