91Ó°ÊÓ

A Binomial Distribution is a common discrete probability distribution. It describes the number of successes in a fixed number of trials, each with the same probability of success. In our exercise, we look at whether adults have sleepwalked.
We consider each adult as a trial. If the adult has sleepwalked, it counts as a success.
The key parameters in a binomial distribution are 'n', the number of trials (or adults surveyed), and 'p', the probability of success (here, the proportion of sleepwalkers).
Using our case:- n = 1480 (total number of surveyed adults)- p = 0.292 (probability that an adult has sleepwalked)
This setup helps us compute probabilities related to our binomial variable.

Normal Approximation is helpful when dealing with binomial distributions, especially with large sample sizes. It makes calculations easier by approximating the binomial distribution with a normal distribution.
In our problem, we used normal approximation because we have a large sample size of 1480 adults.
First, we need the mean (μ) and standard deviation (σ) of our binomial distribution, calculated as follows:- Mean (μ) = n * p = 1480 * 0.292 = 432.16- Standard deviation (σ) = \(\text{√}(n * p * (1 - p))\) = \(\text{√}(1480 * 0.292 * (1 - 0.292))\) ≈ 17.81
Using this normal distribution approximation, we can find probabilities for different numbers of sleepwalkers.

The Z-score helps us determine how far a value is from the mean, measured in standard deviations. It's a crucial step in using the normal approximation to the binomial distribution.
Using the formula:
\[ Z = \frac{X - μ}{σ} \]
where X is the observed number of successes.
For our case with X = 455,Z = \(\frac{455 - 432.16}{17.81}\) ≈ 1.282
This tells us that 455 is 1.282 standard deviations above the mean of our approximated normal distribution. We use this Z-score to find the probability in the next step.

Once we have our Z-score, we look up its corresponding probability in a Z-table (or use a standard normal distribution calculator).
A Z-score of 1.282 corresponds to a probability of 0.10 (or 10%). This is the probability of getting a Z-score less than 1.282.
Since we are interested in 455 or more sleepwalkers, we need the probability of getting a Z-score greater than or equal to 1.282, which is also 10% for this case.
This probability tells us how likely it is to observe such an extreme number of sleepwalkers, given the assumed rate of 29.2%.

Significance Testing helps decide if observed results are unusual enough to suggest that something other than chance is at work. We often use a significance level of 0.05 (5%).
In our case, the probability calculated (0.10) is compared to the significance level:
- If the probability is less than 0.05, the result is considered significantly high.
- Here, the probability (0.10) is greater than 0.05, so the result isn't significantly high.
This means our observed value of 455 sleepwalkers is plausible under the assumption that 29.2% of adults sleepwalk. It doesn’t provide strong evidence to challenge this assumed rate.