The number of viewers of American Idol has a mean of 33 million with a standard deviation of 10 million. Assume this distribution follows a normal distribution.
What is the probability that next week's show will:
a) Have between 37 and 44 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
(b) Have at least 29 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
(c) Exceed 48 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
The number of viewers of American Idol has a mean of 33 million with a standard...
A normal population has a mean of 19 and a standard deviation of 5. a. Compute the z value associated with 22. (Round your answer to 2 decimal places.) Z 0.60| b. What proportion of the population is between 19 and 22? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Proportion c. What proportion of the population is less than 14? (Round z-score computation to 2 decimal places and your final answer to...
A normal population has a mean of 18 and a standard deviation of 5. a. Compute the z value associated with 24. (Round your answer to 2 decimal places.) 2 1.20 b. What proportion of the population is between 18 and 24? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) Proportion c. What proportion of the population is less than 13? (Round z-score computation to 2 decimal places and your final answer to...
A normal population has a mean of 11.2 and a standard deviation of 3.4. Refer to the table in Appendix B.1. a. Compute the z-value associated with 14.3. (Round the final answer to 2 decimal places.) z = b. What proportion of the population is between 11.2 and 14.3? (Round z-score computation to 2 decimal places and the final answer to 4 decimal places.) Proportion c. What proportion of the population is less than 10.0? (Round z-score computation to 2...
A normal population has a mean of 18.3 and a standard deviation of 5. Refer to the table in Appendix B.1 a. Compute the z-value associated with 25.0. (Round the final answer to 2 decimal places.) z = b. What proportion of the population is between 18.3 and 25.0? (Round z-score computation to 2 decimal places and the final answer to 4 decimal places.) Proportion c. What proportion of the population is less than 18.0? (Round z-score computation to 2...
Homework Exercise 7-33 (LO7-4) Dottie's Tax Service specializes in federal tax returns for professional clients, such as physicians, dentists, accountants, and lawyers. A recent audit by the IRS of the returns she prepared indicated that an error was made on 7% of the returns she prepared last year. Assuming this rate continues into this year and she prepares 80 returns, what is the probability that she makes errors on a. More than 6 returns? (Round your z-score computation to 2...
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Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $135,000. This distribution follows the normal distribution with a standard deviation of $42,000. If we select a random sample of 76 households, what is the standard error of the mean? (Round your answer to the nearest whole number.) What is the expected shape of the distribution of the sample mean? What is the likelihood of selecting a sample with...
A normal population has a mean of 21.0 and a standard deviation of 5.0. a.) Compute the z value associated with 24.0 c.) What proportion of the population is less than 18.0? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $165,000. This distribution follows the normal distribution with a standard deviation of $40,000. If we select a random sample of 50 households, what is the standard error of the mean? (Round your answer to the nearest whole number.) What is the expected shape of the distribution of the sample mean? What is the likelihood of selecting a sample with...
A normal distribution has a mean of 60 and a standard deviation of 10. Refer to the table in Appendix B.1. Determine the value above which 80 percent of the values will occur. (Round z-score computation to 2 decimal places and the final answer to 2 decimal places.) X