Question

The incubation time for hummingbird eggs is approximately normal and has a mean of 16 days and standard deviation of 2 days. Let x = the length of hatching times find p(1518days)

Answer 1

0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 16 days and standard deviation of 2 days.

This means that
`\mu = 16, \sigma = 2`

Probability that the length of hatching times is between 15 and 18 days.

This is the p-value of Z when X = 18 subtracted by the p-value of Z when X = 15. So

X = 18

`Z = (X - \mu)/(\sigma)`

`Z = (18 - 16)/(2)`

`Z = 1`

`Z = 1` has a p-value of 0.8413

X = 15

`Z = (X - \mu)/(\sigma)`

`Z = (15 - 16)/(2)`

`Z = -0.5`

`Z = -0.5` has a p-value of 0.3085

0.8413 - 0.3085 = 0.5328

0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.