Let Z be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. $P(-c le Z le c) = 0.9643$ Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.
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Since the probability is symmetric around the mean (0), we can look for the z-score that corresponds to the area of 0.98215 (0.9643 / 2 + 0.5) in the standard normal distribution table. Looking at the table, we find that the closest value to 0.98215 is 0.9821, Show more…
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