Questions: Complete the statement. Round your answer to the nearest thousandth. In a population that is normally distributed with mean 29 and standard deviation 3, the top 31% of the values are those greater than .

Solution Steps

To find the value above which the top \(31\%\) of the values lie, we first need to calculate the z-score corresponding to the \(69\)th percentile of the normal distribution. The z-score is given by:

\[ z = 0.4959 \]

Next, we convert the z-score to the actual value in the distribution using the formula:

\[ X = \mu + z \cdot \sigma \]

where:

• \(\mu = 29\) (mean)
• \(\sigma = 3\) (standard deviation)

Substituting the values, we have:

\[ X = 29 + 0.4959 \cdot 3 \]

Calculating this gives:

\[ X = 29 + 1.4877 = 30.4877 \]

Rounding to the nearest thousandth, we find:

\[ X \approx 30.488 \]

Final Answer