91Ó°ÊÓ

Waiting in Line A Wendy's manager performed a study to determine a probability distribution for the number of people, \(X\), waiting in line during lunch. The results were as follows: $$ \begin{array}{cc|cc} x & \boldsymbol{P}(\boldsymbol{x}) & \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) \\ \hline 0 & 0.011 & 7 & 0.098 \\ \hline 1 & 0.035 & 8 & 0.063 \\ \hline 2 & 0.089 & 9 & 0.035 \\ \hline 3 & 0.150 & 10 & 0.019 \\ \hline 4 & 0.186 & 11 & 0.004 \\ \hline 5 & 0.172 & 12 & 0.006 \\ \hline 6 & 0.132 & & \\ \hline \end{array} $$ (a) Verify that this is a discrete probability distribution. (b) Draw a graph of the probability distribution. Describe the shape of the distribution. (c) Compute and interpret the mean of the random variable \(X\). (d) Compute the standard deviation of the random variable \(X\). (e) What is the probability that eight people are waiting in line for lunch? (f) What is the probability that 10 or more people are waiting in line for lunch? Would this be unusual?

Step by step solution

01

Verify the probability distribution

Check that the sum of all probabilities equals 1. Add the probabilities: \[0.011 + 0.035 + 0.089 + 0.150 + 0.186 + 0.172 + 0.132 + 0.098 + 0.063 + 0.035 + 0.019 + 0.004 + 0.006 = 1.000 \]Since the sum is 1, this is a valid discrete probability distribution.

02

Draw a graph of the probability distribution

Create a bar graph with the x-axis representing the number of people (X) and the y-axis representing the probabilities P(X). For each value of X, draw a bar up to the height of P(X).

03

Describe the shape of the distribution

Observe the bar graph. The distribution is unimodal and slightly right-skewed, meaning there are fewer high probability values on the right side.

04

Compute the mean (expected value) of X

Use the formula for the mean of a discrete probability distribution: \[E(X) = \sum_{x} [x \cdot P(x)] \]Calculate each product and sum them: \[E(X) = (0 \cdot 0.011) + (1 \cdot 0.035) + (2 \cdot 0.089) + (3 \cdot 0.150) + (4 \cdot 0.186) + (5 \cdot 0.172) + (6 \cdot 0.132) + (7 \cdot 0.098) + (8 \cdot 0.063) + (9 \cdot 0.035) + (10 \cdot 0.019) + (11 \cdot 0.004) + (12 \cdot 0.006) = 4.458 \]The mean is 4.458.

05

Compute the standard deviation of X

Use the formula for the standard deviation of a discrete probability distribution: \[\sigma_X = \sqrt{\sum_{x} [(x - E(X))^2 \cdot P(x)]} \]Calculate each squared difference, then weight by the probabilities and sum: \[Var(X) = ((0 - 4.458)^2 \cdot 0.011) + ((1 - 4.458)^2 \cdot 0.035) + ... + ((12 - 4.458)^2 \cdot 0.006) = 6.224 \]Finally, take the square root of the variance: \[\sigma_X = \sqrt{6.224} = 2.496 \]The standard deviation is 2.496.

06

Probability of eight people waiting

Read the probability directly from the table: \[P(8) = 0.063 \]The probability that eight people are waiting in line is 0.063.

07

Probability of 10 or more people waiting

Sum the probabilities of 10, 11, and 12 people waiting: \[P(X \ge 10) = P(10) + P(11) + P(12) = 0.019 + 0.004 + 0.006 = 0.029 \]The probability that 10 or more people are waiting in line is 0.029.

08

Determine if having 10 or more is unusual

An event is usually considered unusual if its probability is less than 0.05. Since \[P(X \ge 10) = 0.029 \]which is less than 0.05, it would be considered unusual.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Distribution

A probability distribution lists all possible outcomes of a random variable and the corresponding probabilities. In this case, the Wendy's manager studied the number of people, denoted as X, waiting in line. Each probability, P(X), represents how likely that number of people will be waiting.
To verify this is a valid discrete probability distribution, the sum of these probabilities must equal 1. In this exercise, adding up the probabilities for all values of X (0 through 12) indeed gives us 1.
The distribution can also be visualized using a bar graph where the x-axis shows the number of people waiting and the y-axis shows the probabilities. Observing this graph can help describe the distribution's shape. For instance, in this case, the distribution is unimodal (having one peak) and slightly right-skewed, meaning higher values have lower probabilities.

Mean of Random Variables

The mean (or expected value) of a random variable gives us the long-term average if we repeat the experiment many times. It's calculated by multiplying each value of X by its probability and summing these products. For the given exercise, the formula used is:
\[E(X) = \sum_{x} \[x \cdot P(x)\] \]
Applying this formula to our data:
\[E(X) = (0 \cdot 0.011) + (1 \cdot 0.035) + \ldots + (12 \cdot 0.006) = 4.458\]
This means on average, around 4.458 people are expected to wait in line during lunch at Wendy's.

Standard Deviation

The standard deviation measures the spread or variability of a set of values. In probability distributions, it tells us how much the values are expected to vary from the mean. The formula for the standard deviation is:
\[\sigma_X = \sqrt{\sum_{x} \[ (x - E(X))^2 \cdot P(x)\] } \]
First, we calculate the variance using the squared differences from the mean, weighted by the probabilities. For our exercise, the variance \(Var(X)\) is found to be 6.224. Taking the square root of this variance gives us the standard deviation:
\[\sigma_X = \sqrt{6.224} = 2.496\]
This standard deviation of approximately 2.496 shows how much the number of people waiting in line at Wendy’s typically deviates from the average.