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$123.74856 rounded to the nearest cent would be?
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Using the t-distribution, as we have the standard deviation for the sample, we could claim that the mean weight in pounds is in the interval (79.1, 85.7).
<h3>What is a t-distribution confidence interval?</h3>The confidence interval is:
In which:
is the sample mean.
- t is the critical value.
- n is the sample size.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed <em>99% confidence interval</em>, with 100 - 1 = <em>99 df</em>, is t = 2.6259.
The other parameters have values given by:
.
We could claim that the mean is any value in the 99% confidence interval, which has bounds given by:
More can be learned about the t-distribution at brainly.com/question/16162795