Question

86.$1$in $6$wins As a special promotion for its $20$-ounce bottles of soda, a soft drink company printed a message on the inside of each cap. Some of the caps said, “Please try again,” while others said, “You’re a winner!” The company advertised the promotion with the slogan “$1$in$6$wins a prize.” Suppose the company is telling the truth and that every $20$-ounce
bottle of soda it fills has a$1$-in-$6$chance of being a winner. Seven friends each buy one $20$-ounce bottle of the soda at a local convenience store. Let $X=$ the number who win a prize.
(a) Explain why $X$ is a binomial random variable.
(b) Find the mean and standard deviation of $X$. Interpret each value in context.

(c) The store clerk is surprised when three of the friends win a prize. Is this group of friends just lucky, or is the company’s $1$-in-$6$ claim inaccurate? Compute $P(X\ge 3)$ and use the result to justify your answer.

Short answer

(a) The variable $X$ is a binomial random variable.

(b) The mean and standard deviation of $X$are$1.167$and $0.9860$.

(c) $P(X\ge 3)\approx 0.0958=9.58\%$, the probability is more than $5\%$, the result is not surprising.

Step-by-step solution

Part (a) Step 1: Given information 

Seven friends each buy one $20$-ounce bottle of the soda at a local convenience store. Let $X=$the number who win a prize.

Part (a) Step 2: Explanation 

Probability of success $\left(p\right)=\frac{1}{6}$
Number of events $\left(n\right)=7$
The random variable $X$being binomial random variable:

• There are two types of accomplishments: successes and failures.
• Candies are unrelated to one another.
• There is a set number of buddies available.
• In addition, the chance of success is fixed.

Because all of the requirements for a binomial distribution have been established.

Part (b) Step 1: Given information

The mean and standard deviation of $X$ to interpret the value in context.

Part (b) Step 2: Explanation 

The mean and standard deviation is:
$$\begin{gathered} \mu =np \\ =7\left(\frac{1}{6}\right) \\ =1.167 \end{gathered}$$
$$\begin{gathered} \sigma =\sqrt{np(1-p)} \\ =\sqrt{7\left(\frac{1}{6}\right)\left(1-\frac{1}{6}\right)} \\ =0.9860 \end{gathered}$$

Part (c) Step 1: Given information 

The company’s $1$-in-$6$ claim inaccurate. Compute the value for $P(X\ge 3)$.

Part (c) Step 2: Explanation 

Determine the value of$P(X\ge 3)$ by:
$$\begin{gathered} P(X\ge 3)=1-P(X<3) \\ =1^{77}{C}_r\times {\left(\frac{1}{6}\right)}^r\times {\left(1-\frac{1}{6}\right)}^{7-r} \\ =0.0958 \\ =9.58\% \end{gathered}$$