Problem 16 - Basic Statistics for Business & Economics

Question

A telemarketer makes six phone calls per hour and is able to make a sale on \(30 \%\) of these contacts. During the next 2 hours, find: a. The probability of making exactly four sales. b. The probability of making no sales. c. The probability of making exactly two sales. d. The mean number of sales in the 2 -hour period.

Step-by-Step Solution

Answer

a) 0.2311, b) 0.0139, c) 0.2311, d) 3.6

Step 1: Understand the Problem Context

We know the salesperson makes 6 calls per hour and has a 30% chance of making a sale during any call. We need to find probabilities unique to a two-hour period, hence the telemarketer makes a total of 12 calls (6 calls/hour * 2 hours).

Step 2: Determine the Distribution Type

This scenario involves a fixed number of independent trials (phone calls), where each trial results in a sale or not a sale (binary outcome). Thus, it's suited for a binomial distribution with number of trials \(n = 12\) and probability of success \(p = 0.3\).

Step 3: Calculate probability for making exactly 4 sales (Part a)

Use the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(k\) is the number of successes (sales). For exactly 4 sales, \(k = 4\). Thus: \[\binom{12}{4} (0.3)^4 (0.7)^8 \approx 0.2311\]

Step 4: Calculate probability for making no sales (Part b)

Substitute \(k = 0\) in the binomial formula, so: \[P(X = 0) = \binom{12}{0} (0.3)^0 (0.7)^{12} \approx 0.0139\]

Step 5: Calculate probability for making exactly 2 sales (Part c)

Substitute \(k = 2\) in the binomial formula, so: \[P(X = 2) = \binom{12}{2} (0.3)^2 (0.7)^{10} \approx 0.2311\]

Step 6: Compute the Mean Number of Sales (Part d)

The mean of a binomial distribution is given by \(\mu = np\). Thus, the mean number of sales in 2 hours is \(\mu = 12 \times 0.3 = 3.6\).

Key Concepts

Probability Calculation

Probability is a measure that describes the likelihood of a particular event occurring. When working with the binomial distribution, we're often interested in the probability of achieving a specific number of 'successes' out of a set number of 'trials.' In our exercise, the trials are the phone calls, and a success is making a sale.

To calculate probabilities in a binomial distribution, we use the formula:

• \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

where:

• \( n \) is the number of trials (phone calls, which are 12 in our case),
• \( k \) is the number of successful trials (sales),
• \( p \) is the probability of success on a given trial (0.3 for each call), and
• \( 1-p \) is the probability of failure (not making a sale in this case).

The different forms of this formula can then be plugged in with the appropriate values to find out the probability of achieving exactly 4, 0, or 2 sales. Each calculation provides insights into how likely each scenario is.

Mean of Binomial Distribution

The mean of a binomial distribution tells us the average number of successes we would expect over a set number of trials. It's an essential measure because it gives us a central value around which the distribution of outcomes is centered. In terms of sales for the telemarketer, it implies the average number of sales per two-hour period.

The formula to find the mean of a binomial distribution is:

• \( \mu = np \)

where:

• \( n \) is the total number of trials (again, that's 12 phone calls), and
• \( p \) is the probability of making a sale on a single call (0.3 in this scenario).

Following this formula for our example, we calculate \( \mu = 12 \times 0.3 = 3.6 \). This means, on average, we expect the telemarketer to make about 3.6 sales during the 2-hour span. This doesn't necessarily mean they will always make exactly 3 or 4 sales - it's an average over many such periods.

Independent Trials

At the heart of the binomial distribution is the concept of independent trials. Independence here means that the outcome of one trial doesn't affect the outcome of another. For the telemarketer making phone sales, it assumes that each call is unrelated to others.

Why is independence important? It helps in maintaining the integrity of our probability calculations. If each trial were influenced by the previous ones, the probabilities and outcomes would need adjusting.

In mathematical terms, independence means the probability of one call resulting in a sale remains \( p = 0.3 \) regardless of how many previous sales happened. This property allows us to apply the binomial distribution confidently, knowing each call's outcome is unaffected by the telemarketer's previous success or failure.