Math Problem Statement

The quality control manager of a cookie company is inspecting a batch of​ chocolate-chip cookies that has just been baked. If the production process is in​ control, the mean number of chip parts per cookie is 9.09.0. How many cookies in a batch of 100 should the manager expect to discard if the company policy requires that all​ chocolate-chip cookies sold have at least threethree ​chocolate-chip parts? Question content area bottom Part 1 The manager should expect to discard enter your response here cookies. ​(Round to one decimal place as​ needed.)

Solution

To solve this problem, we need to determine the expected number of cookies that have fewer than three chocolate-chip parts and will therefore be discarded.

Given:

  • Mean number of chocolate-chip parts per cookie, μ=9.0\mu = 9.0
  • Batch size, n=100n = 100 cookies
  • Minimum acceptable number of chocolate-chip parts per cookie = 3

Since we're dealing with a distribution of chocolate-chip parts in cookies, and the mean number of chip parts is 9.0, we assume a Poisson distribution is a good fit because it's commonly used for counting events over a fixed interval, such as the number of chocolate-chip parts in a cookie.

The probability mass function (PMF) of a Poisson distribution is given by:

P(X=k)=eμμkk!P(X = k) = \frac{e^{-\mu} \cdot \mu^k}{k!}

where XX is the number of chip parts in a cookie, μ=9.0\mu = 9.0, and kk is the number of chip parts (in this case, k<3k < 3).

Let's calculate the probabilities for X=0X = 0, X=1X = 1, and X=2X = 2:

  1. For X=0X = 0: P(X=0)=e9900!=e9P(X = 0) = \frac{e^{-9} \cdot 9^0}{0!} = e^{-9}

  2. For X=1X = 1: P(X=1)=e9911!=9e9P(X = 1) = \frac{e^{-9} \cdot 9^1}{1!} = 9e^{-9}

  3. For X=2X = 2: P(X=2)=e9922!=81e92P(X = 2) = \frac{e^{-9} \cdot 9^2}{2!} = \frac{81e^{-9}}{2}

Now, sum these probabilities to find the total probability that a cookie has fewer than three chip parts:

P(X<3)=P(X=0)+P(X=1)+P(X=2)P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

Finally, multiply this probability by the total number of cookies (100) to find the expected number of cookies to be discarded.

I'll calculate these values next.The manager should expect to discard approximately 0.6 cookies in a batch of 100, based on the company policy that all cookies sold must have at least three chocolate-chip parts.

This means the probability of a cookie having fewer than three chips is quite low.

Would you like further details on how these probabilities were calculated or have any questions?

Here are 5 related questions you might find useful:

  1. How is the Poisson distribution used in quality control?
  2. What are the assumptions underlying the use of a Poisson distribution in this context?
  3. How would the expected number of discarded cookies change if the mean number of chips per cookie increased?
  4. What if the batch size was larger, say 1000 cookies?
  5. Can this approach be applied to other types of quality control processes?

Tip: Understanding the conditions under which a Poisson distribution applies can help in determining whether it’s the appropriate model for your data.

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Math Problem Analysis

Mathematical Concepts

Probability
Poisson Distribution

Formulas

Poisson distribution PMF

Theorems

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Suitable Grade Level

High School