Given a Normal Distribution with a mean (\(\mu\)) of 65 mg/100ml and a standard deviation (\(\sigma\)) of 25 mg/100ml, we can find the proportions requested in parts (a) and (b) by calculating the corresponding z-scores and using the standard normal distribution table (or a calculator).
(a) Proportion of the population greater than 85 mg/100ml:
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Calculate the z-score for 85 mg/100ml: \[ z = \frac{X - \mu}{\sigma} = \frac{85 - 65}{25} = \frac{20}{25} = 0.8 \]
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Look up the z-score in the standard normal distribution table or use a calculator to find the cumulative probability:
- The cumulative probability for \(z = 0.8\) is approximately 0.7881.
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To find the proportion greater than 85 mg/100ml: \[ P(X > 85) = 1 - P(Z < 0.8) = 1 - 0.7881 = 0.2119 \]
Answer for (a):
The proportion of the population that has a sucrose concentration greater than 85 mg/100ml is approximately 0.2119 or 21.19%.
(b) Proportion of the population less than 45 mg/100ml:
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Calculate the z-score for 45 mg/100ml: \[ z = \frac{X - \mu}{\sigma} = \frac{45 - 65}{25} = \frac{-20}{25} = -0.8 \]
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Look up the z-score in the standard normal distribution table or use a calculator to find the cumulative probability:
- The cumulative probability for \(z = -0.8\) is approximately 0.2119.
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To find the proportion less than 45 mg/100ml: \[ P(X < 45) = P(Z < -0.8) \approx 0.2119 \]
Answer for (b):
The proportion of the population that has a sucrose concentration less than 45 mg/100ml is approximately 0.2119 or 21.19%.