The length of Super Bowl commercials is said to be Normally distributed with a mean of 36 seconds and standard deviation of 8 seconds.
a. Find the probability that a randomly selected Super Bowl commercial is shorter than 45 seconds? (2 points)
b. Find the probability that a randomly selected Super Bowl commercial is between 30 and 50 seconds long? (3 points)
c. Budweiser commercials always seem to fall into the longest 4% of Super Bowl commercials. That means that Budweiser commercials are at least how long? (3 points)
David Lane's webpage is perfect for this type of question and replaces the charts and tables found in the back of textbooks
http://davidmlane.com/hyperstat/z_table.html
Just fill in the data
for c) remember that 96% would be shorter.
You might want to use the second feature of this calculator.
To solve these questions, we'll use the properties of the normal distribution. The normal distribution is a continuous probability distribution often used to model real-world phenomena.
a. To find the probability that a randomly selected Super Bowl commercial is shorter than 45 seconds, we need to calculate the area under the normal curve to the left of 45 seconds.
Step 1: Standardize the value using the z-score formula:
z = (x - μ) / σ
where x is the value we want to find the probability for (45 seconds), μ is the mean (36 seconds), and σ is the standard deviation (8 seconds).
Plugging in the values, we get:
z = (45 - 36) / 8 = 1.125
Step 2: Use the standard normal table or a calculator to find the cumulative probability corresponding to the z-score of 1.125. The cumulative probability represents the area under the curve to the left of the z-score.
Using a standard normal table or a calculator, we find that the cumulative probability for a z-score of 1.125 is approximately 0.8686.
So, the probability that a randomly selected Super Bowl commercial is shorter than 45 seconds is approximately 0.8686 or 86.86%.
b. To find the probability that a randomly selected Super Bowl commercial is between 30 and 50 seconds long, we need to calculate the area under the normal curve between those two values.
Step 1: Standardize the lower value (30 seconds) and the upper value (50 seconds) using the z-score formula:
z1 = (30 - 36) / 8 = -0.75
z2 = (50 - 36) / 8 = 1.75
Step 2: Use the standard normal table or a calculator to find the cumulative probabilities for each z-score individually.
Using a standard normal table or a calculator, we find that the cumulative probability for a z-score of -0.75 is approximately 0.2266 and for a z-score of 1.75 is approximately 0.9599.
Step 3: Calculate the difference between the two cumulative probabilities to find the probability between the two values:
0.9599 - 0.2266 = 0.7333
So, the probability that a randomly selected Super Bowl commercial is between 30 and 50 seconds long is approximately 0.7333 or 73.33%.
c. The statement "Budweiser commercials always seem to fall into the longest 4% of Super Bowl commercials" can be interpreted as finding the cutoff value for the longest 4% of the distribution. We need to find the value that separates the longest 4% from the rest of the distribution.
Step 1: Convert the 4th percentile (longest 4%) to a z-score. The z-score corresponding to a given percentile can be found using the standard normal table or a calculator.
Using a standard normal table or a calculator, we find that the z-score corresponding to the 4th percentile is approximately -1.75.
Step 2: Use the z-score formula to find the corresponding value:
z = (x - μ) / σ
Plugging in the values, we have:
-1.75 = (x - 36) / 8
Step 3: Solve the equation for the unknown value x:
-1.75 * 8 = x - 36
-14 = x - 36
x = 36 - 14
x = 22
So, Budweiser commercials must be at least 22 seconds long to fall into the longest 4% of Super Bowl commercials.