Today, U.S. box office ticket prices are on average $8.00 with variance approximately $4.00. The distribution appears to be normal. (Round probabilities to four decimals and dollar answers to two decimals.) a) What's the probability a ticket will cost more than $12.50? b) What's the probability that the ticket price is between $6.50 and $7.50? c) 60% of the tickets cost above what price? $ d) What's the 88th percentile of this distribution? $
Added by Montserrat C.
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In this case, the variance is $4.00, so the standard deviation is $\sqrt{4.00} = 2.00$. Now, we can use the z-score formula to find the z-scores for each of the given prices. The z-score formula is: $z = \frac{x - \mu}{\sigma}$ where $x$ is the value, $\mu$ is Show moreā¦
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