Questions: The average prison sentence for a convicted felon is 15 years. Suppose the distribution of sentences is approximately normal with a standard deviation of 2.1 years. Find the probability that a randomly selected prison sentence is greater than 18 years. Use Table E and enter your answer correct to four decimal places.

Solution Steps

We need to find the probability that a randomly selected prison sentence \( X \) is greater than 18 years, given that the average prison sentence \( \mu \) is 15 years and the standard deviation \( \sigma \) is 2.1 years. This can be expressed mathematically as:

\[ P(X > 18) \]

To find this probability, we first convert the value of 18 years into a Z-score using the formula:

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

Substituting the values:

\[ Z_{start} = \frac{18 - 15}{2.1} = \frac{3}{2.1} \approx 1.4286 \]

The probability \( P(X > 18) \) can be expressed in terms of the cumulative distribution function \( \Phi \):

\[ P(X > 18) = 1 - P(X \leq 18) = 1 - \Phi(Z_{start}) \]

Using the properties of the normal distribution, we have:

\[ P(X > 18) = \Phi(inf) - \Phi(1.4286) \]

From the output, we find:

\[ P(X > 18) = 0.0766 \]

Final Answer