91Ó°ÊÓ

In this exercise, we deal with a probability distribution related to a binomial scenario. A binomial distribution represents the probability of achieving a number of successes in a fixed number of trials, with a consistent probability of success in each trial.

Here, the likelihood that a person believes in the product improvement (considered as a 'success' in this context) is given as 0.25 or 25%. Conversely, the probability that a person does not believe in it (a 'failure') is 0.75 or 75%.

The sample involves eight individuals, so the trials number, denoted as \( n \), equals 8. For each possible outcome \( r \) ranging from 0 (nobody believes) to 8 (everyone believes), we calculate probabilities using the formula:

• \( P(r) = \binom{8}{r} (0.25)^r (0.75)^{8-r} \).

This expression uses combinations \( \binom{n}{r} \) to represent the different ways \( r \) people might believe in the improvement among eight.

Once these probabilities are computed, we can better understand the likelihood of various numbers of people being convinced by the sales talk.

The mean and standard deviation are essential statistics that help summarize the probability distribution. They provide insights into the expected number of successes and the variability in data respectively.

For a binomial distribution, the mean \( \mu \) reflects the average or expected number of successes, calculated as:

• \( \mu = n \times p \)

In this context, with \( n = 8 \) and \( p = 0.25 \), the mean is:

• \( \mu = 8 \times 0.25 = 2 \)

Thus, on average, two out of eight people will believe in the product’s improvements.

The standard deviation \( \sigma \) measures the spread of the distribution, calculated through:

• \( \sigma = \sqrt{n \times p \times q} \)

Substituting the known values, the standard deviation is:

• \( \sigma = \sqrt{8 \times 0.25 \times 0.75} \approx 1.225 \)

This indicates how much variation from the mean we can expect among the observed number of people who believe in the improvement.

Histograms offer a visual representation of probability distributions, making it easier to understand and interpret the data.

After calculating the probability for each possible outcome \( r = 0, 1, 2, \ldots , 8 \), these values are plotted as a histogram. The x-axis in this case represents the number of people out of eight who believe the product is improved, while the y-axis stands for their corresponding probabilities \( P(r) \).

The height of each bar in the histogram reflects the probability of that particular outcome. The shape and spread of the histogram allow you to quickly gauge the most likely outcomes and understand how the probabilities are distributed across different numbers of successes. It's important to notice where the histogram peaks and the direction it trails off to identify the trend or skewness in belief. This graphical format facilitates the assessment of how effectively the marketing message convinces potential consumers.

To ensure the desired level of certainty in outcomes, a calculation of sample size is crucial, particularly when trying to be confident in an event's occurrence.

In this exercise, we want to ascertain the smallest sample size required to be 99% confident that at least one individual will affirm the product's improvement. We start with the complement probability: the chance that nobody believes the product is improved.

That probability is computed as:

• \( P(r = 0) = (0.75)^n \)

To achieve 99% certainty that this event does not happen (meaning \( r \geq 1 \)), we need \( P(r=0) = 0.01 \), leading to the equation:

• \((0.75)^n = 0.01\)

Solving:

• \( n \log(0.75) = \log(0.01) \)
• \( n \approx \frac{\log(0.01)}{\log(0.75)} \approx 17.87 \)

Thus, a minimum of 18 people is necessary for this level of confidence in the marketing study. This ensures that we are highly likely to see at least one success, offering a powerful indication of the product's perceived improvement.