更改

So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.

So when (n+1)p-1 is an integer, then (n+1)p-1 and (n+1)p is a mode. In the case that (n+1)p-1\notin \Z, then only \lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor is a mode.

所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin \Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref>

所以当<math>(n+1)p-1</math>是一个整数时，<math>(n+1)p-1</math>和<math>(n+1)p</math>是一个模。在<math>(n+1)p-1\notin Z</math>的情况下，只有<math>\lfloor (n+1)p-1\rfloor+1=\lfloor (n+1)p\rfloor</math>是模。<ref>See also {{cite web |first=André |last=Nicolas |title=Finding mode in Binomial distribution |work=[[Stack Exchange]] |date=January 7, 2019 |url=https://math.stackexchange.com/q/117940 }}</ref>