Skip to main content
Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.8.64
Chapter 8, Problem 8.8.64

64. (Use of Tech) 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 < b) is given by the integral:
(1/(22√(2π))) ∫[a to b] e^(-((x-110)/22)²/2) dx.
What percentage of U.S. movies are between 1 hr and 1.5 hr long (60-90 min)?

Verified step by step guidance
1
Step 1: Recognize that the problem involves calculating the percentage of movies within a certain range using the normal distribution formula. The integral provided represents the probability density function for a normal distribution.
Step 2: Convert the given time range (60 to 90 minutes) into the integral bounds 'a' and 'b'. Here, 'a' = 60 and 'b' = 90.
Step 3: Standardize the bounds using the z-score formula: z = (x - μ) / σ, where μ is the mean (110 minutes) and σ is the standard deviation (22 minutes). Compute the z-scores for both bounds: z₁ = (60 - 110) / 22 and z₂ = (90 - 110) / 22.
Step 4: Use a technology tool (such as a graphing calculator or software like Python, MATLAB, or WolframAlpha) to evaluate the integral of the normal distribution function between the standardized z-scores. Alternatively, use the cumulative distribution function (CDF) of the standard normal distribution to find the probabilities corresponding to z₁ and z₂, and subtract P(z₁) from P(z₂).
Step 5: Multiply the resulting probability by 100 to convert it into a percentage. This percentage represents the fraction of U.S. movies with lengths between 60 and 90 minutes.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
12m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, depicting that data near the mean are more frequent in occurrence than data far from the mean. It is characterized by its bell-shaped curve, defined by its mean and standard deviation. In this context, the lengths of U.S. movies follow a normal distribution with a mean of 110 minutes and a standard deviation of 22 minutes, which allows for the calculation of probabilities for different ranges of movie lengths.
Recommended video:
06:11
Limits of Rational Functions: Denominator = 0

Standard Normal Variable

A standard normal variable is a normal random variable that has been standardized to have a mean of 0 and a standard deviation of 1. This is achieved through the transformation z = (x - μ) / σ, where μ is the mean and σ is the standard deviation. In the context of the movie lengths, converting the lengths to a standard normal variable allows us to use standard normal distribution tables or software to find probabilities associated with specific ranges of movie lengths.
Recommended video:
07:15
Separation of Variables

Definite Integral

A definite integral calculates the accumulation of quantities, such as area under a curve, between two specified limits. In this case, the integral represents the probability of movie lengths falling between 60 and 90 minutes. Evaluating this integral involves finding the area under the normal distribution curve between these two points, which corresponds to the percentage of movies that fall within this length range.
Recommended video:
05:43
Definition of the Definite Integral
Related Practice
Textbook Question

76–83. Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals.

76. ∫ [cosθ / (sin³θ - 4sinθ)] dθ

1
views
Textbook Question

Visual proof Let F(x)=∫₀ˣ √(a²−t²) dt. The figure shows that F(x)= area of sector OAB+ area of triangle OBC.

a. Use the figure to prove that

F(x) = (a² sin ⁻¹(x/a))/2 + x√(a²−x²)/2

b. Conclude that ∫ √(a²−x²) dx = (a² sin ⁻¹(x/a))/2 + x√(a²−x²)/2 + C.

Textbook Question

87. Surface area Find the area of the surface generated when the curve f(x) = sin x on [0, π/2] is revolved about the x-axis.

Textbook Question

9–40. Integration by parts Evaluate the following integrals using integration by parts.

40. ∫ e^√x dx

1
views
Textbook Question

7–58. Improper integrals Evaluate the following integrals or state that they diverge.

50. ∫ (from 0 to 9) 1/(x - 1)¹ᐟ³ dx

1
views
Textbook Question

7–58. Improper integrals Evaluate the following integrals or state that they diverge.

42. ∫ (from 3 to 4) 1/(x-3)³ᐟ² dx

1
views