Question

According to the article,"Song Dialects and Colonization in the House Finch" (Condor (1975): 407- 422) reported the mean value of song duration for the population of house finches is 1.5 min with a standard deviation of .9 min. Suppose an SRS of 25 finches is selected. How likely is it that the average song duration of the sample will be greater than 1.7 min?

Answer 1

`P(\bar X >1.7) = P(Z> (1.7-1.5)/(0.18)) = P(Z>1.11) = 1-P(Z<1.11) =0.1335`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the song duration of a population, and for this case we know the distribution for X is given by:

`X \sim N(1.5,0.9)`

Where
`\mu=1.5` and
`\sigma=0.9`

Since the distribution for X is normal then we have that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

We select a sample size of n =25 so then the deviation for the sample mean is given by:

`\sigma_(\bar X)= (0.9)/(√(25))= 0.18`

And we can use the z score formula given by:

`z= (\bar X -\mu)/(\sigma_(\bar X))`

We want to find this probability:

`P(\bar X >1.7)`

And using the z score ,the complement rule and the normal standard distribution table or excel we got:

`P(\bar X >1.7) = P(Z> (1.7-1.5)/(0.18)) = P(Z>1.11) = 1-P(Z<1.11) =0.1335`