Question

Male bluethroats have a complex song which is thought to be used to attract female birds. Let x denote the duration of a randomly selected song (in seconds) from a male bluethroat. The authors of research on bluethroat song report the mean song duration is 13.8 seconds and the standard deviation of song durations is 11.8 seconds. The authors also noted that the song length distribution is not normal.

Required:
a. Let = average song duration (in seconds) for a sample of 36 male bluethroat songs. Is this distribution of the sample mean song duration " normally distributed" ?
b. Find the probability that a sample of 36 male bluethroat songs will have a mean duration greater than 12 seconds. Draw, label, and shade a graph to illustrate your result.

Answer 1

a) Sample size larger than 30, so by the Central Limit Theorem, yes.

b) 0.8199 = 81.99% probability that a sample of 36 male bluethroat songs will have a mean duration greater than 12 seconds. The sketch is given at the end.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean song duration is 13.8 seconds and the standard deviation of song durations is 11.8 seconds.

This means that
`\mu = 13.8, \sigma = 11.8`

Sample of 36

This means that
`n = 36, s = (11.8)/(√(36)) = 1.9667`

a. Let = average song duration (in seconds) for a sample of 36 male bluethroat songs. Is this distribution of the sample mean song duration " normally distributed" ?

Sample size larger than 30, so by the Central Limit Theorem, yes.

b. Find the probability that a sample of 36 male bluethroat songs will have a mean duration greater than 12 seconds. Draw, label, and shade a graph to illustrate your result.

This is 1 subtracted by the p-value of Z when X = 12. So

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

By the Central Limit Theorem

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

`Z = (12 - 13.8)/(1.9667)`

`Z = -0.915`

`Z = -0.915` has a p-value of 0.1801.

1 - 0.1801 = 0.8199

0.8199 = 81.99% probability that a sample of 36 male bluethroat songs will have a mean duration greater than 12 seconds.

Sketch: