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The average cost of a 30 -second TV commercial in the Super Bowl was $ 6.5 million with a standard deviation of .5. What is the probability that a sample of 1530 -second commercials costs more than $5 millions?

User JagaSrik
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Final answer:

The probability that a sample of 1530-second commercials costs more than $5 million is approximately 99.865%.

Step-by-step explanation:

To find the probability that a sample of 15 30-second commercials costs more than $5 million, we need to use the z-score formula:

z = (x - μ) / σ

Where x is the value we're interested in, μ is the mean, and σ is the standard deviation. In this case, x = $5 million, μ = $6.5 million, and σ = $0.5 million.

By substituting these values into the formula, we get:

z = ($5 million - $6.5 million) / $0.5 million = -3

Next, we need to find the area to the left of the z-score of -3 using a standard normal distribution table or calculator. This area represents the probability of a sample of commercials costing less than $5 million, so we subtract it from 1 to find the probability of the sample costing more than $5 million.

Finally, we have:

Probability = 1 - P(z < -3)

Using a standard normal distribution table or calculator, we can find that the area to the left of -3 is approximately 0.00135.

Therefore, the probability that a sample of 15 30-second commercials costs more than $5 million is approximately 1 - 0.00135 = 0.99865, or 99.865%.

User Edebill
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