Suppose that the average pop song length in America is 4 minutes with a standard deviation of 1.25 minutes. It is known that song length is normally distributed. Suppose a sample of 25 songs is taken from the population. What is the approximate probability that the average song length will be less than 3.5 minutes? Round to the nearest thousandth.
To find the probability that the average song length will be less than 3.5 minutes, we can use the Central Limit Theorem.
First, we need to calculate the standard error of the mean (SEM), which is the standard deviation of the sample means. The formula for SEM is given by:
SEM = Standard Deviation / √(Sample Size)
In this case, the standard deviation is 1.25 minutes and the sample size is 25. So, we can calculate the SEM as follows:
SEM = 1.25 / √(25)
= 1.25 / 5
= 0.25 minutes
Next, we need to convert the desired song length of 3.5 minutes into a z-score using the following formula:
z = (X - µ) / SEM
Where X is the desired song length, µ is the population mean, and SEM is the standard error of the mean. Plugging in the values:
z = (3.5 - 4) / 0.25
= -0.5 / 0.25
= -2
Now, we need to find the probability associated with the z-score of -2. We can use a standard normal distribution table or a calculator with a normal distribution function to find this probability.
Looking up the z-score of -2 in a standard normal distribution table, we find that the probability is approximately 0.0228.
Therefore, the approximate probability that the average song length will be less than 3.5 minutes is approximately 0.0228 (rounded to the nearest thousandth).
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability