Questions: Suppose that the antenna lengths of woodlice are approximately normally distributed with a mean of 0.22 inches and a standard deviation of 0.05 inches. What proportion of woodlice have antenna lengths that are at most 0.15 inches? Round your answer to at least four decimal places.

Solution Steps

To find the Z-score for the antenna length of \( X = 0.15 \) inches, we use the formula:

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

Substituting the values:

\[ z = \frac{0.15 - 0.22}{0.05} = -1.4 \]

Thus, the Z-score for the value \( 0.15 \) inches is \( z = -1.4 \).

Next, we need to calculate the probability that the antenna lengths are at most \( 0.15 \) inches. This can be expressed as:

\[ P(X \leq 0.15) = \Phi(Z_{end}) - \Phi(Z_{start}) \]

Where \( Z_{end} = -1.4 \) and \( Z_{start} = -\infty \). The cumulative distribution function \( \Phi \) for \( Z_{start} \) at negative infinity is \( 0 \). Therefore, we have:

\[ P(X \leq 0.15) = \Phi(-1.4) - 0 \]

From the standard normal distribution table, we find:

\[ \Phi(-1.4) \approx 0.0808 \]

Thus, the probability that the antenna lengths are at most \( 0.15 \) inches is \( P(X \leq 0.15) = 0.0808 \).

Final Answer