Question

Actual lengths of pregnancy terms for a particular species of mammal are nearly normally distributed about a mean pregnancy length with a standard deviation of 14 days. About what percentage of births would be expected to occur more than 42 days after the mean pregnancy​ length?

Answer 1

Standard deviation = 14 days

If the mean is denoted by , The z score corresponding to 42 days above mean pregnancy length = (Interval between mean and 42 days above mean)/standard deviation

= (+42 - )/14

= 42/14

= 3

P(births would occur more than 42 days after the mean pregnancy​ length) = P(Z > 3)

= 1 - P(Z < 3)

= 1 - 0.9987

= 0.0013

Percentage of births that would be expected to occur more than 42 days after the mean pregnancy​ length = 0.0013 x 100

= 0.13%

Note: (According to empirical rule, 99.7% is within 3 standard deviations of mean. If we use this rule, percentage of births that would be expected to occur more than 42 days after the mean pregnancy​ length = (100 - 99.7)/2 = 0.15%)