To solve the problems, we will use the properties of the normal distribution. Given that the mean (\(\mu\)) is 65 mg/100ml and the standard deviation (\(\sigma\)) is 25 mg/100ml, we can convert the values to z-scores to find the required proportions.
The formula for the z-score is: \[ z = \frac{X - \mu}{\sigma} \] where:
- \(X\) is the value we are looking for,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.
(a) What proportion of the population is greater than 85 mg/100ml?
- Calculate the z-score for \(X = 85\): \[ z = \frac{85 - 65}{25} = \frac{20}{25} = 0.8 \]
- Look up the z-score of 0.8 in the standard normal distribution table or use a calculator:
- The area to the left of \(z = 0.8\) is approximately 0.7881.
- To find the proportion greater than 85 mg/100ml, subtract this value from 1: \[ P(X > 85) = 1 - 0.7881 = 0.2119 \]
So, the proportion of the population that is greater than 85 mg/100ml is approximately 0.2119 or 21.19%.
(b) What proportion of the population is less than 45 mg/100ml?
- Calculate the z-score for \(X = 45\): \[ z = \frac{45 - 65}{25} = \frac{-20}{25} = -0.8 \]
- Look up the z-score of -0.8:
- The area to the left of \(z = -0.8\) is approximately 0.2119.
- The proportion of the population less than 45 mg/100ml is therefore: \[ P(X < 45) = 0.2119 \]
So, the proportion of the population that is less than 45 mg/100ml is approximately 0.2119 or 21.19%.
(c) What proportion of the population lies between 45 and 85 mg/100ml?
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From previous calculations, we know:
- \(P(X < 45) = 0.2119\)
- \(P(X < 85) = 0.7881\)
-
To find the proportion that lies between 45 and 85 mg/100ml, we subtract the two probabilities: \[ P(45 < X < 85) = P(X < 85) - P(X < 45) = 0.7881 - 0.2119 = 0.5762 \]
So, the proportion of the population that lies between 45 and 85 mg/100ml is approximately 0.5762 or 57.62%.