Problem 53 - Basic Statistics for Business and Economics

Question

DeKorte Tele-Marketing, Inc., is considering purchasing a machine that randomly selects and automatically dials telephone numbers. DeKorte Tele- Marketing makes most of its calls during the evening, so calls to business phones are wasted. The manufacturer of the machine claims that their programming reduces the calling to business phones to 15 percent of all calls. To test this claim the Director of Purchasing at DeKorte programmed the machine to select a sample of 150 phone numbers. What is the likelihood that more than 30 of the phone numbers selected are those of businesses, assuming the manufacturer's claim is correct?

Step-by-Step Solution

Answer

The likelihood that more than 30 numbers are businesses is about 3.7%.

Step 1: Identify Given Information

First, we need to identify the information provided in the problem. We know that the probability of dialing a business phone number, according to the manufacturer, is 15%, or \( p = 0.15 \). The sample size is 150 phone numbers, so \( n = 150 \). We are tasked with finding the probability that more than 30 phone numbers are business numbers, which means \( X > 30 \).

Step 2: Define Random Variable and Distribution

The random variable \( X \) represents the number of business phone numbers dialed in a sample of 150 phone numbers. \( X \) follows a binomial distribution because each phone call is a Bernoulli trial with two outcomes: business or not. Hence, \( X \sim \text{Binomial}(n=150, p=0.15) \).

Step 3: Calculate Mean and Standard Deviation

For a binomial distribution, the mean \( \mu \) and standard deviation \( \sigma \) are given by: \[ \mu = np = 150 \times 0.15 = 22.5 \] \[ \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.15 \times 0.85} \approx 4.487 \]

Step 4: Use Normal Approximation

Since \( n \) is large, we can use the normal approximation of the binomial distribution. \( X \sim \mathcal{N}(\mu=22.5, \sigma^2=4.487^2) \). We need to find \( P(X > 30) \). When using the normal approximation, a continuity correction is applied: \( P(X > 30) \approx P(X > 30.5) \).

Step 5: Calculate Z-Score

Convert the problem to a standard normal distribution. Calculate the Z-score for \( X = 30.5 \): \[ Z = \frac{X - \mu}{\sigma} = \frac{30.5 - 22.5}{4.487} \approx 1.784 \]

Step 6: Find Probability Using Z-Score

Use standard normal distribution tables or a calculator to find the probability of \( Z > 1.784 \). \( P(Z > 1.784) \approx 0.037 \). This is the probability that more than 30 phone numbers dialed are business numbers.

Key Concepts

Normal Approximation

When dealing with binomial distributions, if the sample size is large, it becomes cumbersome to calculate probabilities for each possible outcome. Instead, we can use the normal approximation. This is a method where the binomial distribution is approximated by a normal distribution. The basic rule is that if the product of the number of trials, \( n \), and the probability, \( p \), is large enough (usually \( np > 5 \) and \( n(1-p) > 5 \)), you can use a normal approximation.

In the context of DeKorte Tele-Marketing's problem, we have \( n = 150 \) phone numbers and \( p = 0.15 \). The product \( np = 22.5 \) satisfies the condition for normal approximation. By using this approach, probabilities are easier to calculate using the properties of the normal distribution, which is symmetric and defined by a mean \( \mu \) and standard deviation \( \sigma \).

Z-score

The Z-score is a statistical measure that tells us how many standard deviations an element is from the mean. For the purposes of this exercise, Z-scores are used to translate our binomial distribution problem into a standard normal distribution.

To find the Z-score, we use the formula:

• \( Z = \frac{X - \mu}{\sigma} \)

where \( X \) is the value for which we are calculating the probability, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. In our example, \( \mu = 22.5 \) and \( \sigma \approx 4.487 \).

By converting our value \( X = 30.5 \) to a Z-score, we can use the standard normal distribution table to find the probability. The Z-score calculation gives us a sense of how extreme a particular observation is under the assumed conditions.

Probability Calculation

Once you've calculated the Z-score, you can determine the probability by using a standard normal distribution table or a calculator that includes the standard normal cumulative distribution function.

In our exercise, the Z-score was found to be approximately 1.784. The next step is to use a table to find the probability that \( Z > 1.784 \). The table provides the area under the curve to the left of the Z-score, so to find \( P(Z > 1.784) \), you calculate:

• Find \( P(Z < 1.784) \) using the table or calculator.
• Subtract this result from 1 to get \( P(Z > 1.784) \).

This resulting probability shows the likelihood that more than 30 of the dialed numbers in our example are business calls, which is straightforward with the normal approximation.

Continuity Correction

The continuity correction is a crucial step when using the normal approximation for a binomial distribution. Since the binomial distribution is discrete and the normal distribution is continuous, an adjustment is necessary.

This adjustment is applied to better approximate the discrete distribution. When we are looking for the probability that \( X \) is greater than a certain number, we add 0.5 to that number: \( X > k \) becomes \( X > k + 0.5 \). In the DeKorte example, where \( X > 30 \), the correction gives us \( X > 30.5 \). This helps align the binomial distribution with the continuous nature of the normal distribution.

Using continuity correction bridges the gap between the step-like nature of binomial probabilities and the smooth curve of the normal distribution, making calculations more accurate and reliable.