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the results of a fitness trial is a random variable x which is normally distributed with mean µ and standard deviation 2.4. a researcher uses the results from a random sample of 90 trials to calculate a 98% confidence interval for µ. what is the width of this interval ?

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Width of confidence interval is given by
2z_( \alpha /2) ( \sigma )/( √(n) ) = 2 * 2.33 * (2.4)/( √(90) ) =4.91
User Guilherme Oliveira
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