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In probability and statistics, the normal approximation is a technique used to simplify the calculations of binomial probabilities. This approach is particularly useful when dealing with large sample sizes. The binomial distribution, while accurate, can be cumbersome to work with for large values of n (number of trials). Since the binomial distribution can resemble a normal distribution when n is large and p is moderate, we can use the normal (Gaussian) distribution to approximate it. This simplifies calculations significantly.

To use the normal approximation for a binomial distribution, we ensure that both np and n(1-p) are greater than 5. This rule ensures the distribution is adequately symmetric and bell-shaped for the normal distribution to be a good fit.

In our exercise, we approximated the binomially distributed random variable X (students preferring a boy) by a normal distribution with mean \(μ = np\) and standard deviation \(\sigma = \sqrt{np(1−p)}\)\.

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Z-scores are fundamental in statistics and help standardize different data points on a common scale. This scale indicates how many standard deviations an element is from the mean.

The z-score formula is given by:
\[\text{z} = \frac{X − \mu}{\sigma}\]

Here, X is the observed value, \(μ\) is the mean of the distribution, and \(σ\) is the standard deviation.

By converting our binomial variable to a z-score, we can then use the standard normal distribution (z-distribution) to find probabilities more easily.

In the exercise, we found that for X = 75 students preferring a boy, \(\text{z} = \frac{75 − 55.5}{5.91} \approx 3.30\). This means that 75 students preferring a boy is 3.30 standard deviations above the mean expected by the Gallup poll.

With the z-score, we can now reference standard normal distribution tables to find corresponding probabilities. This step drastically simplifies the process of determining how likely or unlikely an observed outcome is.

Statistical significance measures how likely it is that an observed result is due to something other than random chance.

In hypothesis testing, researchers often use a significance level (denoted as α). A common α value is 0.05, representing a 5% risk of concluding that there is an effect when there is no actual effect. If the probability of the observed result, given that the null hypothesis is true, is less than α, we reject the null hypothesis.

In our exercise, the calculated probability of obtaining at least 75 students out of 150 preferring a boy, when the actual population preference is 37%, was found to be very low (0.0005 or 0.05%). This probability is significantly lower than the common α value of 0.05, indicating that such a result is extremely unlikely under the Gallup poll's assumption.

Hence, we consider the result to be statistically significant and conclude that our survey findings likely contradict the Gallup poll.

Always remember, statistical significance does not indicate the size or importance of an effect but rather how confidently we can reject the null hypothesis.