The number of viewers of American Idol has a mean of 21 million with a standard deviation of 10 million. Assume this distribution follows a normal distribution. Refer to the table in Appendix B.1. What is the probability that next week's show will:
(a) Have between 22 and 31 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" (B1?) to find the proportions related to the Z scores.
To find the probability that next week's show will have between 22 and 31 million viewers, we need to use the standard normal distribution table.
First, we need to calculate the z-scores for both 22 million viewers and 31 million viewers. The z-score formula is given by:
z = (x - μ) / σ,
where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
For 22 million viewers:
z1 = (22 - 21) / 10 = 0.1
For 31 million viewers:
z2 = (31 - 21) / 10 = 1.0
Next, we consult the standard normal distribution table (also known as the z-table) to find the corresponding probabilities for these z-scores.
Using the z-table, we find that the probability of getting a z-score of 0.1 is approximately 0.5400, and the probability of getting a z-score of 1.0 is approximately 0.8413.
To find the probability between these two z-scores, we subtract the lower probability from the higher probability:
P(22 ≤ x ≤ 31) = P(z1 ≤ Z ≤ z2) = P(z ≤ 1.0) - P(z ≤ 0.1) = 0.8413 - 0.5400 = 0.3013 (rounded to 4 decimal places).
Therefore, the probability that next week's show will have between 22 and 31 million viewers is approximately 0.3013.