hiya, i've somehow gotten really stuck on this 1-mark q.
X~N(5,16) Write down the value of P(X≠2). [1]
idk why i'm stuck on it but wouldn't it just be 1? idk like i was thinking P(X≠2) would be equal to P(X<2)+P(X>2) but the answer just feels wrong to be honest. any help or clarification would be great, thanks for bearing w me hahaha
hiya, i've somehow gotten really stuck on this 1-mark q.
X~N(5,16) Write down the value of P(X≠2). [1]
idk why i'm stuck on it but wouldn't it just be 1? idk like i was thinking P(X≠2) would be equal to P(X<2)+P(X>2) but the answer just feels wrong to be honest. any help or clarification would be great, thanks for bearing w me hahaha
Your thinking is good! For a continuous distribution X, strict inequalities are treated the same as inclusive ones. So, P(X<2)+P(X>2) = P(X<=2)+P(X>2) = 1