Posted by **Anonymous** on Wednesday, October 1, 2008 at 3:50pm.

The weekly salaries of 100 recent graduates of a private women's college are normally distributed with a mean of $600 and a standard deviation of $80. Determine the interval about the sample mean that has a 1% level of confidence. Use t=2.58

I don't know.

600/sqrt100

60

80+-2.58*60

2334.8 and -74.8

- Math -
**economyst**, Wednesday, October 1, 2008 at 5:34pm
Its 600 +_ 2.58*80

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