To solve the problems, we will use the properties of the normal distribution. Given that the weights of broilers are normally distributed with a mean (μ) of 1446 grams and a standard deviation (σ) of 180 grams, we can calculate the required probabilities using the standard normal distribution (z-scores).
(a) What proportion of broilers weigh between 1185 and 1286 grams?
We first need to find the z-scores corresponding to 1185 grams and 1286 grams.
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For 1185 grams: \[ z = \frac{X - \mu}{\sigma} = \frac{1185 - 1446}{180} = \frac{-261}{180} \approx -1.45 \]
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For 1286 grams: \[ z = \frac{X - \mu}{\sigma} = \frac{1286 - 1446}{180} = \frac{-160}{180} \approx -0.89 \]
Now, we will use a standard normal distribution table (or a calculator) to find the probabilities corresponding to these z-scores.
- For \( z = -1.45 \), the probability \( P(Z < -1.45) \approx 0.0735 \) (using z-table or calculator).
- For \( z = -0.89 \), the probability \( P(Z < -0.89) \approx 0.1867 \).
Now, to find the proportion of broilers that weigh between 1185 and 1286 grams: \[ P(1185 < X < 1286) = P(Z < -0.89) - P(Z < -1.45) \approx 0.1867 - 0.0735 = 0.1132 \]
Thus, the proportion of broilers weighing between 1185 and 1286 grams is approximately 0.1132.
(b) What is the probability that a randomly selected broiler weighs more than 1551 grams?
We will find the z-score for 1551 grams:
\[ z = \frac{X - \mu}{\sigma} = \frac{1551 - 1446}{180} = \frac{105}{180} \approx 0.5833 \]
Now we will find the probability \( P(Z > 0.5833) \), which can be calculated as: \[ P(Z > 0.5833) = 1 - P(Z < 0.5833) \]
Looking up \( z = 0.5833 \) in the standard normal distribution table, we find \( P(Z < 0.5833) \approx 0.7190 \).
Thus, \[ P(Z > 0.5833) \approx 1 - 0.7190 = 0.2810. \]
Therefore, the probability that a randomly selected broiler weighs more than 1551 grams is approximately 0.2810.
(c) Is it unusual for a broiler to weigh more than 1800 grams?
We will find the z-score for 1800 grams:
\[ z = \frac{X - \mu}{\sigma} = \frac{1800 - 1446}{180} = \frac{354}{180} \approx 1.9667 \]
Now we find the probability \( P(Z > 1.9667) \): \[ P(Z > 1.9667) = 1 - P(Z < 1.9667) \]
Looking up \( z = 1.9667 \) in the standard normal distribution table, we find \( P(Z < 1.9667) \approx 0.9750 \).
Thus, \[ P(Z > 1.9667) \approx 1 - 0.9750 = 0.0250. \]
Since the probability of a broiler weighing more than 1800 grams is 0.0250, which is less than 0.05, it is considered unusual.
Therefore, we can fill in the answer: Is it unusual for a broiler to weigh more than 1800 grams? yes, because the probability is 0.0250.
Final Summary of Answers
(a) 0.1132
(b) 0.2810
(c) Yes, because the probability is 0.0250.
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