Answer

Solution

1. 1

   A. A standard result: when $np\in\mathbb{Z}$, the mode equals $np$, which is the mean. True.

2. 2

   B. For $\text{Bin}(n,\tfrac12)$ with $n$ even, the distribution is symmetric about $n/2$, so $n/2$ is the median. True.

3. 3

   C. $\binom{n}{k}p^{k}(1-p)^{n-k}=\binom{n}{k}p^{n-k}(1-p)^{k}$ for all $k$ requires $(p/(1-p))^{k}=(p/(1-p))^{n-k}$ for all $k$, which forces $p=1-p$, i.e.\ $p=\tfrac12$. True.

4. 4

   D. The mode of $\text{Bin}(n,p)$ lies at $\lfloor(n+1)p\rfloor$. When $(n+1)p$ is an integer, BOTH $(n+1)p-1$ and $(n+1)p$ are modes, so the distribution is bi-modal, not unimodal. Hence the statement is incorrect.

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