A normal population has mean U=33 and standard deviation O=7 .

To solve the given problems, we will use the properties of the standard normal distribution. The formula to convert a normal random variable \(X\) to a standard normal variable \(Z\) is given by:

\[ Z = \frac{X - \mu}{\sigma} \]

where \(\mu\) is the mean and \(\sigma\) is the standard deviation.

Given that:

• Mean \(\mu = 33\)
• Standard deviation \(\sigma = 7\)

Part (a): Proportion of the population between 18 and 29

1. Calculate the Z-scores for \(X = 18\) and \(X = 29\):

   For \(X = 18\):

   \[ Z_{18} = \frac{18 - 33}{7} = \frac{-15}{7} \approx -2.1429 \]

   For \(X = 29\):

   \[ Z_{29} = \frac{29 - 33}{7} = \frac{-4}{7} \approx -0.5714 \]

2. Use the Z-table (or a calculator) to find the probabilities corresponding to these Z-scores.

   • The probability for \(Z_{18} \approx -2.1429\) is approximately \(P(Z < -2.14) \approx 0.0162\).
   • The probability for \(Z_{29} \approx -0.5714\) is approximately \(P(Z < -0.57) \approx 0.2843\).

3. To find the proportion of the population between \(18\) and \(29\), subtract the two probabilities:

   \[ P(18 < X < 29) = P(Z < -0.5714) - P(Z < -2.1429) \approx 0.2843 - 0.0162 = 0.2681 \]

Part (b): Probability that a randomly chosen value will be between 26 and 36

1. Calculate the Z-scores for \(X = 26\) and \(X = 36\):

   For \(X = 26\):

   \[ Z_{26} = \frac{26 - 33}{7} = \frac{-7}{7} = -1 \]

   For \(X = 36\):

   \[ Z_{36} = \frac{36 - 33}{7} = \frac{3}{7} \approx 0.4286 \]

2. Find the probabilities corresponding to these Z-scores.

   • The probability for \(Z_{26} = -1\) is \(P(Z < -1) \approx 0.1587\).
   • The probability for \(Z_{36} \approx 0.4286\) is \(P(Z < 0.43) \approx 0.6664\).

3. To find the probability that a randomly chosen value will be between \(26\) and \(36\), subtract the two probabilities:

   \[ P(26 < X < 36) = P(Z < 0.4286) - P(Z < -1) \approx 0.6664 - 0.1587 = 0.5077 \]

Final Answers:

(a) The proportion of the population between 18 and 29 is approximately \(0.2681\).

(b) The probability that a randomly chosen value will be between 26 and 36 is approximately \(0.5077\).

Thus, rounded answers:

• (a) 0.2681
• (b) 0.5077