Question

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 ?

Answer 1

Width of confidence interval is given by $2z_{ \alpha /2} \frac{ \sigma }{ \sqrt{n} } = 2 \times 2.33 \times \frac{2.4}{ \sqrt{90} } =4.91$