91Ó°ÊÓ

Probability is a fundamental concept in statistics and mathematics. It tells us how likely an event is to occur. In our exercise, the probability is the likelihood that any given automobile is not insured. This probability is given as 23%, or in decimal form, 0.23.

• A probability of 0 means an event will not happen.
• A probability of 1 means an event is certain to occur.
• Probabilities must always lie between 0 and 1.

Understanding probability helps us determine expected outcomes and analyze risks in various scenarios.

The expected value is a crucial concept in statistics that represents the average or mean outcome of a random event if it were repeated many times. It's often used in business statistics to anticipate future outcomes.For our specific problem, the expected number of uninsured automobiles can be found using the formula:\[ E = n \times p \]where:

• \( n \) is the total number of automobiles, which is 35 in our case.
• \( p \) is the probability of an automobile being uninsured, 0.23 here.

Substituting these values into the equation, we find that the expected number of uninsured automobiles is 8.05. This means, on average, we expect about 8 cars to be uninsured if we had many such events.

Variance assesses how much the outcomes of a probability distribution differ from the expected value. It indicates the spread or variability in a set of data points, helping us understand the risk or uncertainty involved.For a binomial distribution, the variance is computed using the formula:\[ \text{Var} = n \times p \times (1 - p) \]In our scenario:

• \( n = 35 \), the total number of trials (automobiles).
• \( p = 0.23 \), the probability of being uninsured.

Calculating gives us a variance of 6.1975. This value reflects the degree of variation we can expect from the average number of uninsured cars in such a group.

The standard deviation is a measure of the amount of variation or dispersion in a set of values. It is the square root of the variance and provides an easily interpretable value for variation, as it is in the same unit as the original data.For the binomial distribution problem of uninsured automobiles, the standard deviation is:\[ \text{SD} = \sqrt{\text{Var}} \]Inserting our variance of 6.1975, we find:\[ \text{SD} = \sqrt{6.1975} \approx 2.49 \]This shows the average deviation from the mean number of uninsured cars. A smaller standard deviation would imply that the number of uninsured cars is consistently near the average (8.05). In contrast, a larger value would indicate more significant variability.

The binomial distribution represents the probability of a fixed number of successes in a given number of independent trials, each with the same probability of success. Here, the ‘success’ is an automobile not being insured. Key characteristics of a binomial distribution:

• There is a fixed number of trials: 35 automobiles.
• Each trial has two possible outcomes: insured or not insured.
• The probability of success remains constant, here 0.23.
• Each trial is independent, meaning the outcome does not affect others.

Using the binomial distribution helps us model and compute the probabilities associated with various outcomes in our example.