Question

Normal distribution of movie lengths A study revealed that the lengths of U.S. movies are normally distributed with a mean of 110 minutes and a standard deviation of 22 minutes. This means that the fraction of movies with lengths between \(a\) and \(b\) minutes (with \(a

Short answer

Answer: To find the percentage, calculate the integral of the normal distribution function with the limits of 60 minutes and 90 minutes, and then multiply the result by 100 to get the percentage.

Step-by-step solution

Convert the time range into minutes

Since the given range is in hours, we need to convert that range into minutes to match the given mean (110 minutes) and standard deviation (22 minutes) for US movie lengths distribution. 1 hour = 60 minutes, and 1.5 hours = 90 minutes.

Substitute the given mean, standard deviation, and limits to the integral formula

Now, we have a = 60 minutes and b = 90 minutes. So the integral becomes: $$ \frac{1}{22 \sqrt{2 \pi}} \int_{60}^{90} e^{-((x-110)/22)^{2}/2} dx $$

Calculate the integral

Calculating the integral, we get the fraction of movie lengths between 60 and 90 minutes: $$ \text{Fraction} =\frac{1}{22 \sqrt{2 \pi}} \int_{60}^{90} e^{-((x-110)/22)^{2}/2} dx $$

Convert the result into a percentage

To get the percentage of movies between those lengths, we need to multiply the result of the integral by 100: $$ \text{Percentage} = \text{Fraction} \times 100 $$ Now you have the percentage of US movies with lengths between 1 hour (60 minutes) and 1.5 hours (90 minutes).

Key concepts

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion within a set of values. In the context of a normal distribution, which is a type of probability distribution, the standard deviation shows how much each value in the data set varies from the mean (average), formally known as the expected value. A higher standard deviation indicates that the data points spread out more widely from the mean, showing a higher degree of variation.

Regarding the exercise on movie lengths, the standard deviation of 22 minutes means that the typical deviation of movie lengths from the average (110 minutes) is 22 minutes. This knowledge is crucial because it sets the scale for our probability calculations: we can predict that most movies will fall within one standard deviation (88-132 minutes) from the mean, and this range contains about 68% of the data if it follows a normal distribution. To visualize the significance of standard deviation imagine a plot where the horizontal axis shows the length of movies and the curve's width represents the standard deviation.

Probability Density Function

The probability density function (PDF) describes the likelihood of a random variable taking on specific values. For a continuous variable, such as movie lengths in our case, the PDF gives the probability of the variable falling within a particular range of values instead of exact points. The normal distribution has a bell-shaped PDF, symmetrical about the mean.

The function for the normal distribution is given by the following equation:
\[ f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \]
where \(\mu\) is the mean, and \(\sigma\) is the standard deviation. In the movie length example, the PDF indicates the probability that any randomly selected movie will have a certain length. The specified PDF equation in the problem, once the mean and standard deviation are substituted, reflects the particular normal distribution of movie lengths.

Definite Integrals

Definite integrals are a fundamental concept in calculus, providing a way to calculate the accumulation of quantities, like the probability of continuous random variables falling between two points in the context of the exercise. In probability theory, the integral over the probability density function between two points gives the total probability of falling within that interval. The formula for the area under the curve of the PDF between two points \(a\) and \(b\) is:
\[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \]
where \(f(x)\) is the PDF and \(P(a \leq X \leq b)\) is the probability that \(X\), a random variable, falls between \(a\) and \(b\). In the example of movie lengths, the definite integral equation provided calculates the exact proportion of movies that are between 60 and 90 minutes long. To find the percentage of movies in that length range, we calculate the integral value and convert it into a percentage as demonstrated in the solution steps.