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farmer noticed that some of the eggs laid by his hens had double yolks. He stimated the probability of this happening to be 0.05 . Eggs are packed in boxes of 12 . Find the probability that in a box, the number of eggs with double yolks will be exactly one. A customer bought three boxes. Find the probability that only 2 of the boxes contained exactly 1 egg with double yolk. The farmer delivered 10 boxes to a local shop. Using a suitable approximation, find the probability that the delivery contained at least 9 eggs with double yolks. (6 marks) The weight of an individual egg can be modelled by a normal distribution with mean 65 g and standard deviation 2.4 g . Find the probability that a randomly chosen egg weighs more than 68 g .

Ask by Mcfarlane Zimmerman. in Malaysia
Mar 19,2025

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1. In a box of 12 eggs, the probability of exactly one egg having double yolks is calculated using the binomial probability formula. 2. For a customer buying 3 boxes, the probability that exactly 2 boxes contain exactly one egg with double yolks is determined using the binomial distribution. 3. When the farmer delivers 10 boxes (120 eggs), the probability of having at least 9 eggs with double yolks is approximated using the Poisson distribution. 4. The probability that a randomly chosen egg weighs more than 68 grams is found by standardizing the normal distribution and using standard normal tables.

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To solve the first part, we can use the binomial probability formula \( P(X = k) = C(n, k) p^k (1-p)^{n-k} \), where \( n = 12 \), \( k = 1 \), and \( p = 0.05 \). For exactly one double-yolked egg: \[ P(X = 1) = C(12, 1) (0.05)^1 (0.95)^{11} \approx 0.3867 \] For the second part, we need to find the probability that exactly 2 out of 3 boxes contain exactly 1 egg with a double yolk. We can again use the binomial formula with \( n = 3 \), \( k = 2 \), and \( p \) as the previously calculated probability of a box containing exactly 1 double-yolked egg: \[ P(Y = 2) = C(3, 2) (P(X=1))^2 (1-P(X=1))^1 \approx 0.1841 \] For the last part, applying a suitable normal approximation for the box that the farmer delivered, we can assume the number of eggs with double yolks is binomial with \( n = 120 \) (10 boxes, 12 eggs each) and \( p = 0.05 \). We calculate the mean \( \mu = np = 6 \) and standard deviation \( \sigma = \sqrt{np(1-p)} \approx 4.86 \). Using the normal approximation: We find \( P(Z \geq 9) \) or \( P(X \geq 9) \) using \( Z = \frac{X - \mu}{\sigma} \). Now, for the weight of an egg, we wish to find the probability that an egg weighs more than 68g: Using the standard normal distribution: \[ Z = \frac{68 - 65}{2.4} \approx 1.25 \] Consulting a Z-table gives us \( P(Z > 1.25) \approx 0.1056 \). --- Did you know that double-yolked eggs are often associated with younger hens? These hens are more prone to laying eggs with two yolks due to their hormonal development. This interesting quirk makes these eggs a delightful surprise for breakfast! For further reading, dive into statistical modeling and probabilities! A good recommendation is "Statistics for Dummies," which makes concepts like binomial probabilities and normal distributions more approachable. You'll be crunching numbers like a pro in no time!

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