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

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

You're correct.

The idea is, with an example, it is impossible to find someone that's exactly 170cm tall, not 170.1cm, not 169.9cm.

Alternatively, in a continuous distribution, calculus tells us $P(X \neq 2) = 1-P(X=2) = 1-\int^{2}_{2}{P(X=x)}dx = 1$.

This does not work in discrete.

The idea is, with an example, it is impossible to find someone that's exactly 170cm tall, not 170.1cm, not 169.9cm.

Alternatively, in a continuous distribution, calculus tells us $P(X \neq 2) = 1-P(X=2) = 1-\int^{2}_{2}{P(X=x)}dx = 1$.

This does not work in discrete.

(edited 9 months ago)

Original post by aditi_idk

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

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

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