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
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
1 answer
1 answer