Questions: A normal population has mean μ=9 and standard deviation σ=5. (a) What proportion of the population is less than 20 ? (b) What is the probability that a randomly chosen value will be greater than 6 ? Round the answers to four decimal places. The proportion of the population less than 20 is 0.9861. The probability that a randomly chosen value will be greater than 6 is 0.2743.

Solution Steps

To find the proportion of the population that is less than 20, we calculate the cumulative distribution function (CDF) at \( x = 20 \). The Z-score is calculated as follows:

\[ Z_{end} = \frac{20 - \mu}{\sigma} = \frac{20 - 9}{5} = 2.2 \]

Using the properties of the normal distribution, we find:

\[ P(X < 20) = \Phi(Z_{end}) - \Phi(-\infty) = \Phi(2.2) - 0 = 0.9861 \]

Thus, the proportion of the population less than 20 is:

\[ \boxed{0.9861} \]

To find the probability that a randomly chosen value will be greater than 6, we first calculate the Z-score for \( x = 6 \):

\[ Z_{start} = \frac{6 - \mu}{\sigma} = \frac{6 - 9}{5} = -0.6 \]

The probability is then calculated as:

\[ P(X > 6) = 1 - P(X < 6) = 1 - \Phi(Z_{start}) = 1 - \Phi(-0.6) \]

Using the properties of the normal distribution, we find:

\[ P(X > 6) = \Phi(\infty) - \Phi(-0.6) = 1 - 0.2743 = 0.7257 \]

Thus, the probability that a randomly chosen value will be greater than 6 is:

\[ \boxed{0.2743} \]

Final Answer