Confused about variance formula in Central Limit Theorem - Discrete Random Variable

Hello everyone,

I have a question about the variance formula presented in the video on the Central Limit Theorem for discrete random variables. Why is the variance \sigma^2 equal to np(1-p)?

P.S. I used the formula Var(X) = \Bbb{E}[X^2] - \Bbb{E}[X]^2 to verify it, and it matches the result of np(1-p), but I don’t quite understand why it is np(1-p).

Thank you!

Central Limit Theorem - Discrete Random Variable | Coursera

Hello @zhengyiwang82

Number of Trials (n): More trials generally lead to greater variability.
Probability of Success (p): When p is close to 0 or 1, the outcome is more predictable (less variance). When p is around 0.5, the outcome is less predictable (more variance).
Failure Probability (1−p): The product p(1−p) represents the interaction between success and failure probabilities.

Thus, the variance np(1−p) encapsulates the inherent variability in a binomial process, depending on both the number of trials and the probabilities of success and failure.

Hope this helps

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Hello @jenitta.jebaraj

Thank you for your response. I apologize for not describing my question clearly. I understand the factors that contribute to variance, but I am still unclear about why the variance formula for the binomial distribution is specifically np(1-p).

My main confusion lies in understanding why the variance in this context is np(1-p) instead of the general formula we learned earlier, Var(X) = \Bbb{E}[X^2] - \Bbb{E}[X]^2.

Could you please explain the derivation or reasoning behind why the variance for a binomial distribution is np(1-p), and how it relates to the general variance formula?