Asked by Anonymous
Suppose heights of men are normally distributed with a mean 69.0 inches and standard deviation 2.8 inches. What percent of men are over 6 feet tall? Round to the nearest tenth of a percent.
I understanf finding the z score, but I'm not sure how to find the percent.
I understanf finding the z score, but I'm not sure how to find the percent.
Answers
Answered by
John
Find the z-score
72-69 divided by 2.8 = 1.07 for a z-score.
What percent are over 6 feet tall, so we want the right tail.
You can use a z-table to find the value. Unfortunately, not all z-tables are constructed the same way. If you have a TI-83 or 84 calculator, you can use the DISTR function found above the VARS key.
2nd distr Normalcdf
you enter (1.07, 1000000000000) use a very large number to take you all the way out to the right tail. I got .1423 that means 14.23% would be taller than 6 feet.
If you are using a z-table, compare your answer with what the calculator gives.
72-69 divided by 2.8 = 1.07 for a z-score.
What percent are over 6 feet tall, so we want the right tail.
You can use a z-table to find the value. Unfortunately, not all z-tables are constructed the same way. If you have a TI-83 or 84 calculator, you can use the DISTR function found above the VARS key.
2nd distr Normalcdf
you enter (1.07, 1000000000000) use a very large number to take you all the way out to the right tail. I got .1423 that means 14.23% would be taller than 6 feet.
If you are using a z-table, compare your answer with what the calculator gives.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.