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 less than 0.23 inches? Round your answer to at least four decimal places.

Answer 1

Solution:

Given: X =  the antenna lengths of woodlice are approximately normally distributed with a mean of 0.2 inches and a standard deviation of 0.05 inches.

That is:

We have to find:

P( woodlice have antenna lengths that are less than 0.23 inches) =.............?

P( X < 0.23) = .............?

Find z score:

Thus we get:

P(X < 0.23) = P( Z < 0.60)

Look in z table for z = 0.6 and 0.00 and find area,

P( Z < 0.60) = 0.7257

Thus:

P(X < 0.23) = P( Z < 0.60)

P(X < 0.23) = 0.7257

Thus the proportion of woodlice have antenna lengths that are less than 0.23 inches is 0.7257