Questions: A normal population has mean μ=9 and standard deviation σ=6. Find the proportion of the population that is greater than 6. Round the answers to at least four decimal places. The proportion of the population that is greater than 6 is.

Solution Steps

To find the proportion of the population that is greater than 6, we need to calculate the z-score for the value 6 using the given mean and standard deviation. Then, we use the z-score to find the cumulative probability from the standard normal distribution table. Finally, subtract this cumulative probability from 1 to get the proportion of the population that is greater than 6.

To find the proportion of the population that is greater than 6, we first calculate the z-score using the formula: \[ z = \frac{x - \mu}{\sigma} \] where \( x = 6 \), \( \mu = 9 \), and \( \sigma = 6 \). Substituting the values, we get: \[ z = \frac{6 - 9}{6} = -0.5 \]

Using the z-score, we find the cumulative probability from the standard normal distribution. The cumulative probability for \( z = -0.5 \) is approximately: \[ P(Z \leq -0.5) = 0.3085 \]

The proportion of the population that is greater than 6 is given by: \[ P(X > 6) = 1 - P(Z \leq -0.5) = 1 - 0.3085 = 0.6915 \]

Final Answer