80 U2 songs (in seconds) were selected at random. 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(mean=289.7 or 290, standard deviation= 76.8 or 77). 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 are 90% of all U2 songs expected to be?
Thanks in advance

To calculate the percentages, we need to use the z-score formula in conjunction with the normal distribution table.

A.) To find the percentage of U2 songs that are expected to be over 180 seconds, we first need to convert this value to a z-score using the formula:

z = (x - μ) / σ

Where x is the value (180 seconds), μ is the mean (290 seconds), and σ is the standard deviation (77 seconds).

z = (180 - 290) / 77 ≈ -1.4286

Now, we can find the corresponding percentage from the normal distribution table. The table provides the percentage of values that fall below a given z-score, so we subtract this percentage from 100 to find the percentage of values above the z-score.

Looking up the z-score of -1.43 in the table, we find that the percentage below this value is approximately 0.0764. Subtracting this from 1 gives us:

Percentage above 180 seconds = 1 - 0.0764 ≈ 0.9236

So, approximately 92.36% of U2 songs are expected to be over 180 seconds.

B.) To find the percentage of U2 songs that are expected to be between 210 seconds and 300 seconds, we need to calculate the z-scores for both values.

z1 = (210 - 290) / 77 ≈ -1.039

z2 = (300 - 290) / 77 ≈ 0.130

We can now look up these z-scores in the normal distribution table and find the corresponding percentages.

Percentage between 210 seconds and 300 seconds = Percentage below z2 - Percentage below z1

Looking up the z-scores, we find that the percentage below -1.039 is approximately 0.1469 and the percentage below 0.13 is approximately 0.5528.

Therefore, the percentage between 210 seconds and 300 seconds is:

Percentage between 210 seconds and 300 seconds = 0.5528 - 0.1469 ≈ 0.4059

So, approximately 40.59% of U2 songs are expected to be between 210 seconds and 300 seconds.

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 for the cumulative percentage of 90%.

Looking up the cumulative percentage of 90% in the normal distribution table, we find that the z-score is approximately 1.2816.

We can now use the z-score formula to solve for the length of time (x).

z = (x - μ) / σ

1.2816 = (x - 290) / 77

Rearranging the equation and solving for x, we get:

x = (1.2816 * 77) + 290 ≈ 387.6

Therefore, approximately 90% of all U2 songs are expected to be under 387.6 seconds.

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http://davidmlane.com/hyperstat/z_table.html