Question

The mean duration of television commercials on a given network is 75 seconds, with a standard deviation of 20 seconds. Assume that durations are approximately normally distributed. a. What is the approximate probability that a commercial will last less than 35 seconds? b. What is the approximate probability that a commercial will last longer than 55 seconds?

Short answer

Answer: a) The approximate probability that a commercial will last less than 35 seconds is 0.0228. b) The approximate probability that a commercial will last longer than 55 seconds is 0.8413.

Step-by-step solution

Write down the given information

The mean duration (µ) of television commercials on a given network is 75 seconds, and the standard deviation (σ) is 20 seconds.

Calculate z-scores

For both (a) and (b), we need to find the z-scores using the formula: z = (X - µ) / σ - For part (a), X = 35 seconds z = (35 - 75) / 20 z = -40 / 20 z = -2 - For part (b), X = 55 seconds z = (55 - 75) / 20 z = -20 / 20 z = -1

Find probabilities using z-scores

Now we will find the probability of each scenario using the z-score and a standard normal distribution table or a calculator. - For part (a), we need to find the probability that z < -2. Using a standard normal distribution table or calculator, we find the probability: P(z < -2) ≈ 0.0228 - For part (b), we need to find the probability that z < -1. Using a standard normal distribution table or calculator, we find the probability: P(z < -1) ≈ 0.1587

Answer the questions

Finally, we can answer the questions in the exercise. a. The approximate probability that a commercial will last less than 35 seconds is 0.0228. b. To find the probability that a commercial will last longer than 55 seconds, we can subtract the probability found in part (b) from 1 because the total probability equals 1 in a normal distribution: 1 - P(z < -1) = 1 - 0.1587 ≈ 0.8413. Therefore, the approximate probability that a commercial will last longer than 55 seconds is 0.8413.

Key concepts

z-score

The z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It tells us how many standard deviations the value is from the mean. This is very helpful when dealing with a normal distribution, as it allows us to determine how likely or unlikely that value is to occur.

To calculate the z-score, use the formula:

• \( z = \frac{X - \mu}{\sigma} \)

Here, \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

For example, if a commercial lasts 35 seconds, and the average duration is 75 seconds with a standard deviation of 20 seconds, the z-score is:

• \( z = \frac{35 - 75}{20} = -2 \)

So, the commercial lasts 2 standard deviations less than the average duration. This makes it an unlikely event in this distribution.

probability

Probability in statistics refers to the likelihood or chance of a particular outcome occurring. When values follow a normal distribution, we can find the probability of a given value by calculating its z-score and referring to standard normal distribution tables.

For instance, if you want to find the probability that a commercial lasts less than 35 seconds, knowing its z-score is -2, you consult the table and find that the probability (P(z < -2)) is 0.0228.

• This means there is a 2.28% chance of a commercial being shorter than 35 seconds.

This process helps answer such questions by providing precise probabilities for normally distributed data.

standard_deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation implies that most of the numbers are close to the mean, while a high standard deviation indicates that the numbers are more spread out.

In our example, the standard deviation of commercial durations is 20 seconds. This tells us that the durations typically vary by 20 seconds from the average duration of 75 seconds.

• In contexts like these, the standard deviation helps us understand the spread of the data and how typical or atypical certain durations might be.

Understanding standard deviation is crucial because it is often used in various statistical formulas and helps describe the properties of a data set in relation to its mean.

mean

The mean, or average, is one of the most common measures in statistics. It helps us summarize a set of values with a single number that represents the central point of the data set.

To compute the mean, add up all the values and divide by the number of values:

• Mean (\( \mu \)) = \( \frac{\text{Sum of all data points}}{\text{Number of data points}} \)

In the given exercise, the mean duration of the commercials is 75 seconds. This means that if you were to watch all commercials, their average length would be 75 seconds.

The mean is particularly useful because it sets a reference point for calculating other important statistics, like the z-score, and helps compare individual data points against the average.