Problem 33 - Calculus for Biology and Medicine

Question

Clementine oranges are sold in boxes. Each box contains 50 clementines. The probability that a clementine in a box is spoiled is \(0.01\) (a) Use an appropriate approximation to determine the probability that a box contains 0,1, or at least 2 spoiled clementines. (b) A shipment of clementines (said to be hybrid crossings between oranges and tangerines) with 100 boxes is considered unacceptable if \(35 \%\) or more of the boxes contain spoiled clementines. What is the probability that a shipment is unacceptable?

Step-by-Step Solution

Answer

The probabilities are: 0 spoiled: 0.6065, 1 spoiled: 0.3033, at least 2 spoiled: 0.0902. The shipment is unacceptable with a probability of 0.8133.

Step 1: Understanding the Binomial Distribution

Each box contains 50 clementines, so we can model the number of spoiled clementines in a box as a binomial random variable, where the number of trials is 50 and the probability of spoilage for each clementine is 0.01. This is noted as \( X \sim\text{Binomial}(n=50, p=0.01) \).

Step 2: Approximation with Poisson Distribution

Due to the small probability of spoilage and a large number of trials, the Poisson distribution can approximate the binomial distribution. The parameter \( \lambda \) of the Poisson distribution is \( n \cdot p = 50 \cdot 0.01 = 0.5 \).

Step 3: Calculation of Probability for 0 Spoiled Clementines

The probability of 0 spoiled clementines is given by \( P(X = 0) \). Using the Poisson distribution, this probability is \( e^{-0.5} \cdot \frac{0.5^0}{0!} = e^{-0.5} \approx 0.6065 \).

Step 4: Probability for 1 Spoiled Clementine

The probability of exactly 1 spoiled clementine is \( P(X = 1) \). Using the Poisson distribution: \( e^{-0.5} \cdot \frac{0.5^1}{1!} = 0.5 \cdot e^{-0.5} \approx 0.3033 \).

Step 5: Calculation for At Least 2 Spoiled Clementines

The probability for at least 2 spoiled clementines is \( P(X \geq 2) \). This can be calculated as \( 1 - P(X = 0) - P(X = 1) = 1 - 0.6065 - 0.3033 \approx 0.0902 \).

Step 6: Understanding the Criteria for an Unacceptable Shipment

The shipment of 100 boxes is unacceptable if more than \( 35\% \) of the boxes have spoiled clementines. This corresponds to at least \( 35 \) boxes out of 100 containing at least 1 spoiled clementine.

Step 7: Approximating the Distribution of Spoiled Boxes

Using the Poisson approximation, the probability that a box contains at least 1 spoiled clementine is \( 1 - P(X = 0) = 1 - 0.6065 = 0.3935 \). For 100 boxes, the number of boxes with spoiled clementines follows a binomial distribution \( Y \sim \text{Binomial}(n=100, p=0.3935) \).

Step 8: Using Normal Approximation for Binomial Distribution

Given \( n \) is large, the normal approximation can be used: \( Y \sim N(100 \cdot 0.3935, 100 \cdot 0.3935 \cdot (1 - 0.3935)) \). The mean is approximately \( 39.35 \) and variance is \( 23.87 \).

Step 9: Calculating Probability of Unacceptable Shipment

Using the normal approximation, standardize \( Y \) and calculate \( P(Y \geq 35) \) by finding \( P(Z \geq \frac{35 - 39.35}{\sqrt{23.87}}) \). This results in \( P(Z \geq -0.89) \approx 0.8133 \).

Key Concepts

Binomial Distribution

The binomial distribution is a key concept in probability theory. It helps us model scenarios where an event can occur with two possible outcomes: success or failure. In the context of clementines, each piece of fruit is either spoiled or not. When dealing with the binomial distribution:

• The number of trials (n) is fixed. Here, it's 50 clementines.
• The probability of success (p), in this case, a clementine being spoiled, is 0.01.
• The trials are independent, meaning one spoiled clementine doesn't affect the others.

We denote this with the notation: \( X \sim \text{Binomial}(n=50, p=0.01) \). The key takeaway is that if you know these parameters, you can determine the probability of any number of spoiled clementines in a box.

Normal Approximation

The normal approximation for a binomial distribution is useful when the number of trials is large. It's employed to simplify probability calculations when direct use of the binomial formula becomes cumbersome. Here's how it works:

• You first ensure that both \( np \) and \( n(1-p) \) are greater than 5, ensuring a good fit.
• The binomial distribution \( \text{Binomial}(n, p) \) can be approximated by a normal distribution \( N(np, np(1-p)) \), where mean is \( np \) and variance is \( np(1-p) \).

For our problem, once we know the probability of a box having spoiled clementines, we can apply the normal approximation to assess if a shipment will be considered unacceptable based on spoilage.

Probability Calculation

Calculating the probability of specific events, like the likelihood of having 0, 1, or more spoiled clementines in a box, involves using probability distributions. Here, the Poisson and normal approximations help streamline these calculations:For example, to determine the probability of having \( X \geq 2 \) spoiled clementines:

• Use the Poisson approximation since \( p \) is small and \( n \) is large, setting \( \lambda = np = 0.5 \).
• Calculate \( P(X = 0) \) and \( P(X = 1) \) and subtract these from 1 gives \( P(X \geq 2) \).

For the shipment, since it contains many boxes, we apply the normal approximation to determine the probability that an unacceptable number of boxes are spoiled. By converting \( Y \) (boxes with spoiled clementines) into a standard normal variable \( Z \), you can find \( P(Y \geq 35) \), indicating whether the shipment is likely to be rejected.