To find the required proportions and probabilities for a normal population with mean \(\mu = 67\) and standard deviation \(\sigma = 16\), we can use the standard normal distribution (Z-distribution).
(a) Proportion of the population greater than 105
First, we need to standardize the value of 105 using the Z-score formula:
\[ Z = \frac{X - \mu}{\sigma} \]
where \(X = 105\), \(\mu = 67\), and \(\sigma = 16\).
Calculating the Z-score:
\[ Z = \frac{105 - 67}{16} = \frac{38}{16} = 2.375 \]
Now, we want to find the proportion of the population that is greater than 105, which is \(P(X > 105)\) or \(P(Z > 2.375)\).
Using a standard normal distribution table or calculator, we can find \(P(Z < 2.375)\) first, and then use the complement rule:
\[ P(Z > 2.375) = 1 - P(Z < 2.375) \]
Looking up \(P(Z < 2.375)\):
- From standard normal tables or a calculator, \(P(Z < 2.375) \approx 0.9911\).
Now calculate the proportion greater than 105:
\[ P(Z > 2.375) = 1 - 0.9911 = 0.0089 \]
Thus, the proportion of the population that is greater than 105 is approximately:
\[ \boxed{0.0089} \]
(b) Probability that a randomly chosen value will be less than 82
We repeat the standardization process for the value of 82:
\[ Z = \frac{X - \mu}{\sigma} = \frac{82 - 67}{16} = \frac{15}{16} = 0.9375 \]
Now we want \(P(X < 82)\) or \(P(Z < 0.9375)\).
Using the standard normal table or a calculator, we find \(P(Z < 0.9375)\):
- From standard normal tables or a calculator, \(P(Z < 0.9375) \approx 0.8249\).
Thus, the probability that a randomly chosen value will be less than 82 is approximately:
\[ \boxed{0.8249} \]
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