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Consider the following statements relate...

Consider the following statements related to a veriable X having a binomial distribution `b_(x)(n,p)`
1. If `p = 1/2`, then the distribution is symmetrical.
2. p remaining constant, P(X = r) increases as n increases. Which of the statements given above is/are correct?

A

1 only

B

2 only

C

Both 1 and 2

D

Neither 1 nor 2

Text Solution

AI Generated Solution

The correct Answer is:
To determine the correctness of the statements related to a variable \( X \) having a binomial distribution \( B(n, p) \), we will analyze each statement step by step. ### Step 1: Analyze the first statement **Statement 1**: If \( p = \frac{1}{2} \), then the distribution is symmetrical. - A binomial distribution \( B(n, p) \) is symmetrical when \( p = \frac{1}{2} \). This is because when \( p = \frac{1}{2} \), the probabilities of success and failure are equal, leading to a symmetric distribution around the mean \( \frac{n}{2} \). - Therefore, **Statement 1 is correct**. ### Step 2: Analyze the second statement **Statement 2**: \( p \) remaining constant, \( P(X = r) \) increases as \( n \) increases. - The probability mass function for a binomial distribution is given by: \[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \] - If \( p \) is constant and we increase \( n \), the term \( \binom{n}{r} \) (which represents the number of ways to choose \( r \) successes from \( n \) trials) increases. This can lead to an increase in \( P(X = r) \) for fixed \( r \) as \( n \) increases. - Thus, **Statement 2 is also correct**. ### Conclusion Both statements are correct. ### Final Answer Both statements given above are correct. ---

To determine the correctness of the statements related to a variable \( X \) having a binomial distribution \( B(n, p) \), we will analyze each statement step by step. ### Step 1: Analyze the first statement **Statement 1**: If \( p = \frac{1}{2} \), then the distribution is symmetrical. - A binomial distribution \( B(n, p) \) is symmetrical when \( p = \frac{1}{2} \). This is because when \( p = \frac{1}{2} \), the probabilities of success and failure are equal, leading to a symmetric distribution around the mean \( \frac{n}{2} \). - Therefore, **Statement 1 is correct**. ...
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