91Ó°ÊÓ

In statistics, we often deal with large sample sizes. In such cases, the binomial distribution can be approximated using a normal distribution. This technique simplifies calculations because the normal distribution has well-established properties and tables.

The key idea is to use a normal distribution with the same mean and standard deviation as the binomial distribution. In the exercise, the mean is calculated as \(\text{mean} = n \times p = 200 \times 0.55 = 110\).The standard deviation is \(\text{std} = \sqrt{n \times p \times (1 - p)} = \sqrt{200 \times 0.55 \times 0.45} \approx 7.03\).So, the binomial distribution \(B(200, 0.55)\) is approximated by a normal distribution \(N(110, 7.03)\).

Why do we use this approximation? Handling large numbers with the normal distribution is easier. Plus, for large sample sizes (n generally greater than 30), the normal distribution closely mirrors the binomial distribution.

When we want to find the probability of a specific outcome in a distribution, we perform a probability calculation. After approximating the binomial distribution with a normal distribution, we need to calculate the probability for a value being 130 or more.

Using the normal distribution, we convert the value of interest (130) to the z-score. The z-score helps us find the probability using standard normal distribution tables.

The z-score is calculated using the formula \(z = \frac{x - \text{mean}}{\text{std}}\). In our exercise, this becomes \(z = \frac{130 - 110}{7.03} \approx 2.85\).

Next, we use the z-score to find the probability from a standard normal table. For a z-score of 2.85, the table gives a probability of 0.9978, which represents the area to the left of the z-score (less than 130). To find the probability of 130 or more, we calculate the complement: \(\text{P}(X \ge 130) = 1 - 0.9978 = 0.0022\). This is the probability of obtaining at least 130 respondents.

The z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It tells you how many standard deviations the value is from the mean. Calculating the z-score is a crucial step in converting your binomial problem to a normal problem.

In our example, assume we are studying the probability of males living at home. We first found the mean (110) and standard deviation (7.03). To calculate the z-score for 130 males, we used the formula:

\(\text{z} = \frac{130 - 110}{7.03} = 2.85\).

What does this mean? A z-score of 2.85 implies that 130 males living at home is 2.85 standard deviations above the mean. Using the standard normal distribution table, we find the probability of this z-score. The higher the z-score, the less likely the event, confirming our result is rare under the assumed probability (55% living at home).

Remember, the z-score gives us a way to understand probabilities in the context of the normal distribution, making it easier to handle large sample sizes effectively.