Question

On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose

that you buy one of his cakes.

a. In words, define the random variable $X$.

b. List the values that $X$may take on.

c. Give the distribution of$X.X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

d. On average, how many pieces of egg shell do you expect to be in the cake?

e. What is the probability that there will not be any pieces of egg shell in the cake?

f. Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not

be any egg shell in any of the cakes?

g. Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why?

Short answer

(a) The random variable $X$has the following definition:

$X=$the number of shell pieces in a single cake.

(b) Because the random variable $X$has a Poisson distribution with an upper bound of $100,$the value is given by $x=0,1,2,\dots$

(c) The result is $X~P(1.5)$.

(d) the result is $\mu =1.5\text{.}$

(e) The result we get $P(x=0)=0.2231\text{.}$

(f) The result $P(x=0)=0.0001.$

(g) The result$P(x=7)=0.00075$.

Step-by-step solution

Explanation (a)

The random variable$X$has the following definition:

$X=$the percentage of seniors who participated in after-school sports throughout their four years of high school.

Explanation (b)

Make a list of all the possible values for X.

Because the random variable $X$has a Poisson distribution with an upper bound of 100, the value is given by $x=0,1,2,\dots$

Explanation (c)

The random variable $X$will be distributed according to the Poisson distribution. In every two cakes, three pieces of egg shell are given, and in each piece of cake, one piece of egg shell is given.

$=\frac{3}{2}$

$=1.5$

$X~P(1.5)$

Explanation (d)

The Poisson distribution predicted is equal to,

$\mu =1.5$

As a result, we you except $1.5$egg shell fragments in the cake.

Explanation (e)

The required probability can be represented as $P(x=0)$, and it can then be determined with a calculator.

As a result, the chance that there would be no fragments of egg shell in the cake is $P(x=0)=0.2231$.

Explanation (f)

It is provided. If there are $3$pieces of egg shell in every $2$cakes, then there are 6 pieces of egg shell in each cake.

$$\begin{gathered} =\frac{3}{2}\times 6 \\ =9 \end{gathered}$$

The random variable $X$will follow Poisson distribution as below,

$X~P\left(9\right)$

The required probability can be expressed as$P(x=0)$ and then determined with a calculator.

$23498041\mathrm{E}-4$

As a result, the probability $P(x=0)=0.0001$is the likelihood that no egg shell will be found in any of the cakes.

Explanation (g)

Probability can be determined using $P(x=7)$rather than a calculator. $7.564261847{\mathrm{E}}^{-4}$

As a result, based on the average supplied for Pierre, probability $P(x=7)=0.00075$is not zero; yes, there could be seven pieces of shell in the cake.