# Math

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

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1. You have reversed your values.

The standard error is the standard deviation — not the mean — divided by the square root of n.

You also want the mean (\$600) [not the standard deviation] ± 2.58 times the standard error.

I hope this helps. Thanks for asking.

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posted by PsyDAG

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