In your question, when you refer to \( p = 0.5 \) and mention a normal distribution, it seems that you are discussing the sample size for a proportion in a statistics context. The proportion \( p \) typically refers to the probability of success in a binomial distribution.
When calculating the sample size for proportions, the formula used is derived from the normal approximation to the binomial distribution when \( n \) is sufficiently large.
The formula to calculate the sample size \( n \) for estimating a proportion is:
\[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \]
Where:
- \( Z \) is the Z-score corresponding to the desired confidence level.
- \( p \) is the estimated proportion (in this case, 0.5).
- \( E \) is the margin of error you are willing to accept.
To complete your statement:
Since you mentioned \( p = 0.5 \), you can say that the distribution you are dealing with is called the binomial distribution.
In the context of proportions, it is understood that for large sample sizes, the binomial distribution can be approximated by a normal distribution, which is why you might see "normal distribution" mentioned in connection with \( p \). However, the underlying discrete distribution is indeed the binomial distribution.
If you provide a desired confidence level and margin of error, I can help you calculate the sample size using the formula mentioned above.
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