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Suppose that, on average, an American professional football game lasts about six hours. Assume that game times are normally distributed with a standard deviation of 1.4 hours: a. Find the probability that a game lasts less than 5.5 hours. (Do not round intermediate calculations. Round your final answer to 4 decimal places.) b. 11% of football game times are below what time? (Do not round intermediate calculations. Round your final answer to 2 decimal places.)

User Wmac
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Final answer:

The probability that a game lasts less than 5.5 hours is approximately 0.3571. 11% of football game times are below 4.58 hours.

Step-by-step explanation:

a. To find the probability that a game lasts less than 5.5 hours, we need to calculate the z-score for the value 5.5 using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. Once we have the z-score, we can use a standard normal distribution table or a calculator to find the probability associated with that z-score. The z-score for 5.5 is z = (5.5 - 6) / 1.4 ≈ -0.3571. Using the standard normal distribution table or a calculator, we find that the probability of a game lasting less than 5.5 hours is approximately 0.3571.

b. To find the time below which 11% of football game times are, we need to find the z-score corresponding to the 11th percentile. This can be done by finding the z-score that corresponds to an area of 0.11 in the lower tail of the distribution. Using a standard normal distribution table or a calculator, we find that the z-score corresponding to an area of 0.11 is approximately -1.224. Using the formula z = (x - μ) / σ, we can solve for x to find the corresponding time. Rearranging the formula, we have x = μ + z * σ, where μ is the mean, σ is the standard deviation, and z is the z-score. Plugging in the values, we get x = 6 + (-1.224) * 1.4 ≈ 4.58. Therefore, 11% of football game times are below 4.58 hours.

User Helpa
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