Posted by **Mary** on Friday, January 23, 2009 at 11:56am.

Th weekly salaries of a sample 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

Not sure if this is correct?

=600+or-2.58(80)/10

=600+or-206.4/10

=600+or-20.64

- Pre-calculus -
**drwls**, Friday, January 23, 2009 at 1:37pm
I am not sure what "t" is supposed to be, but the interval that the salary will be in 99% of the time is 600 + or - 206.4

206.4 happens to be 2.58 times the standard deviation.

Your t may be what other texts usually call s, the deviation from the mean divided by sigma

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