Respuesta :
Answer:
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.
Step-by-step 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = 13.8, \sigma = 11.8[/tex]
Sample of 36
This means that [tex]n = 36, s = \frac{11.8}{\sqrt{36}} = 1.9667[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{12 - 13.8}{1.9667}[/tex]
[tex]Z = -0.915[/tex]
[tex]Z = -0.915[/tex] 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:
