From the length(in seconds) of 80 randomly chosen U2 songs, the mean determined was 289.6986295 and standard deviation was 76.82292002. Assume that the data is normally distributed and that the bands entire collection of songs has a mean and standard deviation equal to those calculated above. What percentage of U2 songs are expected to be
A.) over 180 seconds
B.) Between 210 seconds and 300 seconds
C.) Under what length of time will 90% of all U2 songs expected to be?
To answer these questions, we will use the properties of the normal distribution and calculate the probabilities based on the given mean and standard deviation.
A) To find the percentage of U2 songs expected to be over 180 seconds, we need to find the area under the normal distribution curve to the right of 180 seconds.
Step 1: Calculate the z-score for 180 seconds using the formula:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
z = (180 - 289.6986295) / 76.82292002
z ≈ -1.4316
Step 2: Find the area to the right of the z-score using a standard normal distribution table or a calculator.
The area to the right of -1.4316 can be found as 1 - area to the left of -1.4316.
Using a standard normal distribution table, the area to the left of -1.4316 is approximately 0.0764. Therefore, the area to the right of -1.4316 is 1 - 0.0764 = 0.9236.
So, the percentage of U2 songs expected to be over 180 seconds is approximately 92.36%.
B) To find the percentage of U2 songs expected to be between 210 seconds and 300 seconds, we need to find the area under the normal distribution curve between these two values.
Step 1: Calculate the z-scores for 210 seconds and 300 seconds.
z1 = (210 - 289.6986295) / 76.82292002 ≈ -1.03
z2 = (300 - 289.6986295) / 76.82292002 ≈ 1.34
Step 2: Find the area between these two z-scores using a standard normal distribution table or a calculator.
Using the standard normal distribution table, the area to the left of -1.03 is approximately 0.1515, and the area to the left of 1.34 is approximately 0.9099. Therefore, the area between these two z-scores is approximately 0.9099 - 0.1515 = 0.7584.
So, the percentage of U2 songs expected to be between 210 seconds and 300 seconds is approximately 75.84%.
C) To find the length of time under which 90% of all U2 songs are expected to be, we need to find the corresponding z-score that encloses 90% of the area under the normal distribution curve.
Step 1: Find the z-score that encloses 90% of the area. This z-score is known as the z-score for the 90th percentile.
Using a standard normal distribution table or a calculator, the z-score for the 90th percentile is approximately 1.28.
Step 2: Calculate the length of time corresponding to this z-score using the formula:
x = z * σ + μ
where x is the value, z is the z-score, σ is the standard deviation, and μ is the mean.
x = 1.28 * 76.82292002 + 289.6986295
x ≈ 384.75
So, under 90% of all U2 songs are expected to be under 384.75 seconds (approximately 6 minutes and 24 seconds).
you can get a handle on this stuff at
http://davidmlane.com/hyperstat/z_table.html