The length of country and western songs is normally distributed and has a mean of 200 seconds and a standard deviation of 30 seconds. Find the probability that a random selection of 9 songs will have mean length of 186.30 seconds or less. Assume the distribution of the lengths of the songs is normal
To find the probability that a random selection of 9 songs will have a mean length of 186.30 seconds or less, we need to use the central limit theorem.
The central limit theorem states that the sampling distribution of the sample means approaches a normal distribution as the sample size increases, regardless of the shape of the original population distribution.
In this case, since we are given that the length of country and western songs is normally distributed with a mean of 200 seconds and a standard deviation of 30 seconds, we can use this information to calculate the probability.
First, we need to calculate the standard deviation of the sample means, also known as the standard error. The formula for the standard error is:
Standard Error = Standard Deviation of the population / Sqrt(Sample Size)
In this case, the standard deviation of the population is 30 seconds and the sample size is 9 songs. Therefore, the standard error is:
Standard Error = 30 / Sqrt(9) = 10
Next, we need to calculate the z-score for 186.30 seconds using the formula:
z = (x - mean) / standard error
where x is the desired value (186.30 seconds in this case), mean is the population mean (200 seconds), and standard error is the standard error calculated above.
Using the given values, we get:
z = (186.30 - 200) / 10 = -1.37
Now, we need to find the probability of having a mean length of 186.30 seconds or less. To do this, we look up the z-score in the standard normal distribution table. The table gives the area under the standard normal curve to the left of a given z-score.
Using the table, we find that the probability corresponding to a z-score of -1.37 is approximately 0.0844.
Therefore, the probability that a random selection of 9 songs will have a mean length of 186.30 seconds or less is approximately 0.0844, or 8.44%.
To solve this problem, we need to use the Central Limit Theorem, which states that the distribution of sample means will be approximately normally distributed regardless of the shape of the original population, as long as the sample size is sufficiently large (typically, at least 30).
In this case, we have a normal distribution of country and western song lengths, with a mean (μ) of 200 seconds and a standard deviation (σ) of 30 seconds.
Since we want to find the probability that the mean length of a sample of 9 songs is 186.30 seconds or less, we need to calculate the z-score for this value and then use the z-table to find the corresponding probability.
The z-score formula is given by: z = (x - μ) / (σ / √n), where x is the value we want to find the probability for, μ is the mean, σ is the standard deviation, and n is the sample size.
In this case, x = 186.30 seconds, μ = 200 seconds, σ = 30 seconds, and n = 9.
Plugging these values into the formula, we have: z = (186.30 - 200) / (30 / √9)
= -13.70 / (30 / 3)
= -13.70 / 10
= -1.37
Now, we need to find the corresponding probability using the z-table. Looking up the z-score of -1.37 in the z-table, we find that the probability is approximately 0.0844.
Therefore, the probability that a random selection of 9 songs will have a mean length of 186.30 seconds or less is approximately 0.0844, or 8.44%.