Question

A normal population has mean u=67 and standard deviation o=16.

(a) What proportion of the population is greater than 105?
(b) What is the probability that a randomly chosen value will be less than 82?
Round the answers to four decimal places.

Answer 1

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:

\[
\boxed{0.8249}
\]