Posted by **Maile** on Sunday, February 10, 2013 at 4:31pm.

The body temperatures of adults are normally distributed with a mean of 98.6 degree F and a standard deviation of 0.60 degree F. If 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4 degree F.

- Statistics -
**MathGuru**, Sunday, February 10, 2013 at 5:40pm
Use z-scores:

z = (98.4 - 98.6)/(0.60/√36) = ?

Once you calculate the z-score, check a z-table for your probability. (Remember that the problem is asking to find the probability "greater than" 98.4)

- Statistics -
**Jane**, Sunday, February 10, 2013 at 5:55pm
-2 z=0.0228 1-0.0228 answer 0.9772

## Answer this Question

## Related Questions

- statistics - Assume the body temperatures of healthy adults are normally ...
- statistics - Assume that body temperatures of healthy adults are normally ...
- statistics - Assume that body temperatures of healthy adults are normally ...
- statistics - Assume that body temperatures of healthy adults are normally ...
- statistics - Assume the body temperatures of healthy adults are normally ...
- statistics - assume that adults have IQ scores that are normally distributed ...
- statistics - Assume that body temperatures of healthy adults are normally ...
- Statistics - assume that adults have IQ scores that are normally distributed ...
- stats - This week we practice with Binomial Distribution. You can use Appendix ...
- statistics - Assume the body temperatures of healthy adults are normally ...