Question

Suppose that the antenna lengths of woodlice are approximately normally distributed with a mean of 0.2 inches and a standard deviation of 0.05 inches. What proportion of woodlice have antenna lengths that are at most 0.23 inches? Round your answer to at least four decimal places.

Answer 1

X: Antenna length of a woodlice

X follows normal distribution with mean 0.2 inches and standard deviation 0.05 inches.

Probability that a woodlice have antenna length of at most 0.23 inches = P(X 0.23)

Z-score of 0.23 = (0.23 - 0.2)/0.05 = 0.6

From standard normal tables = P(Z 0.6) = 0.7257

P(X 0.23) = P(Z 0.6) = 0.7257

Probability that a woodlice have antenna length of at most 0.23 inches = P(X 0.23) = 0.7257

proportion of woodlice have antenna lengths that are at most 0.23 inches = Probability that a woodlice have antenna length of at most 0.23 inches =0.7257

Proportion of woodlice have antenna lengths that are at most 0.23 inches = 0.7257