CompNerd2014
New member
- Joined
- Jun 22, 2014
- Messages
- 1
| A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 5.1 minutes and the standard deviation was 0.40 minutes. |
| (a) | What fraction of the calls last between 5.1 and 5.8 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) |
| z=5.8-5.1/.4=1.75 .4955 |
| Fraction of calls |
| (b) | What fraction of the calls last more than 5.8 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) |
| z=5.8-5.1/.4=1.75 so do I subtract the .4955 from .5 since 5.8 is greater than 5.1? |
| Fraction of calls |
| (c) | What fraction of the calls last between 5.8 and 6.5 minutes?(Round z-score computation to 2 decimal places and final answer to 4 decimal places.) |
| z=5.8-5.1/.4=1.75 z=6.5-5.1/.4=3.5 .4989-.3531=.1458 subtract because they are both to the right of the mean? |
| Fraction of calls |
| (d) | What fraction of the calls last between 4.5 and 6.5 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.) |
| z=4.5-5.1/.4=-1.5(.3531) z=6.5-5.1/.4=3.5(.4989) Do I subtract each one for .5 then add? |
| Fraction of calls |
| (e) | As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 3% of the calls. What is this time? (Round z-score computation to 2 decimal places and your final answer to 2 decimal places.) |
| I have no idea what to do here |
| Duration |