Question

To increase sales, a local donut shop began putting an extra donut in some of the boxes. Customers are unaware of which boxes had the extra donut and the shop owner claimed that one in seven boxes had an extra donut. Eight customers each bought one box of donuts at the shop. Let X = the number of customers that bought a box containing an extra donut.A. Is X a binomial random variable? Explain.B. What is the mean and standard deviation of X? Provide an interpretation for each value in context.C. Two of the six customers buy a box of donuts with an extra donut. Is the shop's claim accurate? Compute P(X ≥ 2) and use the result to justify your answer.

Answer 1

Answer:

a) For this case we have the following conditions:

1) Independence between thr random trials

2) Probability of success defined and fixed $p=\frac{1}{7}$

3) A value of n fixed n=8

So then all the assumptions are satisfied to use the binomial distribution

Let X the random variable of interest, the distribution would be given by:

$X \sim Binom(n=8, p=\frac{1}{7}=0.143)$

b) The mean is given by:

$E(X) = np= 8* \frac{1}{7} =1.14$

The variance is given by:

$Var(X) = np(1-p) = 8 *\frac{1}{7} (1-\frac{1}{7})=0.980$

And the deviation is just:

$Sd(X) = \sqrt{0.980}=0.990$

c) If we find the probability using the complement rule we got:

$P(X \geq 2) = 1-P(X$

And if we find the individual probabilities we got

$P(X=0)=(8C0)(0.143)^0 (1-0.143)^{8-0}=0.291$  

$P(X=1)=(8C1)(0.143)^1 (1-0.143)^{8-1}=0.388$  

$P(X \geq 2)= 1- [P(X=0) +P(X=1)]= 1-[0.291+0.388]=0.321$

So on this case makes sense the claim since the probability is higher or equal than the expected with the claim

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

For this case we define the random variable X="the number of customers that bought a box containing an extra donut."

Part a

For this case we have the following conditions:

1) Independence between thr random trials

2) Probability of success defined and fixed $p=\frac{1}{7}$

3) A value of n fixed n=8

So then all the assumptions are satisfied to use the binomial distribution

Let X the random variable of interest, the distribution would be given by:

$X \sim Binom(n=8, p=\frac{1}{7}=0.143)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part b

The mean is given by:

$E(X) = np= 8* \frac{1}{7} =1.14$

The variance is given by:

$Var(X) = np(1-p) = 8 *\frac{1}{7} (1-\frac{1}{7})=0.980$

And the deviation is just:

$Sd(X) = \sqrt{0.980}=0.990$

Part c

Assuming this question :"Two of the eight customers buy a box of donuts with an extra donut. Is the shop's claim accurate? Compute P(X ≥ 2) and use the result to justify your answer." Because we can't change the value of n since if we do this we need to change th distribution.

If we find the probability using the complement rule we got:

$P(X \geq 2) = 1-P(X$

And if we find the individual probabilities we got

$P(X=0)=(8C0)(0.143)^0 (1-0.143)^{8-0}=0.291$  

$P(X=1)=(8C1)(0.143)^1 (1-0.143)^{8-1}=0.388$  

$P(X \geq 2)= 1- [P(X=0) +P(X=1)]= 1-[0.291+0.388]=0.321$

So on this case makes sense the claim since the probability is higher or equal than the expected with the claim