Math Problem Statement

The amount of time that a customer waits on hold at a call center can be modeled by a uniform distribution on the interval from 0 seconds to 120 seconds. (a) Draw a density curve to model the amount of time that a randomly selected customer waits on hold. Be sure to include scales on both axes. (b) About what percent of the time does a customer have to wait between 100 and 120 seconds?

Solution

Let's solve the two parts of the problem provided in the image.

Part (a): Draw a density curve

The problem states that the amount of time a customer waits on hold can be modeled by a uniform distribution on the interval [0,120][0, 120] seconds.

For a uniform distribution, the probability density function (PDF) is constant over the interval. The formula for the PDF of a uniform distribution U(a,b)U(a, b) is:

f(x)=1baforaxbf(x) = \frac{1}{b - a} \quad \text{for} \quad a \leq x \leq b

Here, a=0a = 0 and b=120b = 120, so the density function becomes:

f(x)=11200=1120f(x) = \frac{1}{120 - 0} = \frac{1}{120}

This means the density curve is a horizontal line from x=0x = 0 to x=120x = 120, with a constant height of 1120\frac{1}{120}. On the graph:

  • The x-axis represents the time (in seconds) from 0 to 120.
  • The y-axis represents the probability density, which is 1/1201/120.

Part (b): Percent of the time customer waits between 100 and 120 seconds

To find this percentage, we calculate the probability that the wait time is between 100 and 120 seconds for the uniform distribution U(0,120)U(0, 120). The probability for an interval [x1,x2][x_1, x_2] in a uniform distribution is given by:

P(x1Xx2)=x2x1baP(x_1 \leq X \leq x_2) = \frac{x_2 - x_1}{b - a}

Substituting the values x1=100x_1 = 100, x2=120x_2 = 120, a=0a = 0, and b=120b = 120:

P(100X120)=1201001200=20120=160.1667P(100 \leq X \leq 120) = \frac{120 - 100}{120 - 0} = \frac{20}{120} = \frac{1}{6} \approx 0.1667

This means that about 16.67% of the time, a customer has to wait between 100 and 120 seconds.


Would you like a graph for the density curve or further clarifications?

Here are 5 related questions to consider:

  1. How would the density curve change if the maximum wait time increased to 150 seconds?
  2. What would the density function be if the wait time followed a non-uniform distribution?
  3. How would you calculate the probability of waiting less than 50 seconds?
  4. How do you interpret the height of the density curve in a uniform distribution?
  5. How would the probability change if the interval was between 80 and 100 seconds?

Tip: Uniform distributions are useful for modeling situations where all outcomes in a range are equally likely, such as random wait times within a known time limit.

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Math Problem Analysis

Mathematical Concepts

Probability
Uniform Distribution

Formulas

Probability Density Function (PDF) of Uniform Distribution: f(x) = 1/(b - a) for a ≤ x ≤ b
Probability of an Interval in Uniform Distribution: P(x_1 ≤ X ≤ x_2) = (x_2 - x_1)/(b - a)

Theorems

Properties of Uniform Distribution

Suitable Grade Level

Grades 11-12