# A random variable X is normally distributed with mu=60 and sigma = 3. What is the value of 2 numbers a,b so that P(X=a)=P(X=b). The solution is a=60 and b=65. However, I do not know how to come up with that answer. As far as I understand P(X=a) and P(X=b) have to be both 0 since you always have to give a range e.g. P(a<X). Moreover if I insert the values 60 and 65 in the formula Z=(X−mu)/sigma than I would end up with 0,1.667 and z-scores 0.5, 0.952 respectively.

A random variable X is normally distributed with $\mu =60$ and $\sigma$ = 3. What is the value of 2 numbers a,b so that $P\left(X=a\right)=P\left(X=b\right)$.
The solution is $a=60$ and $b=65$.
However, I do not know how to come up with that answer. As far as I understand $P\left(X=a\right)$ and $P\left(X=b\right)$ have to be both 0 since you always have to give a range e.g. $P\left(a. Moreover if I insert the values 60 and 65 in the formula $Z=\left(X-\mu \right)/\sigma$ than I would end up with 0,1.667 and z-scores 0.5, 0.952 respectively.
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Audrey Russell
Your reasoning is valid -- asking for $P\left(X=a\right)=P\left(X=b\right)$ makes essentially no sense because such a probability is always 0 for a continuous distribution.
Indeed, this seems to be the only way to make $a=60,b=65$ a valid answer.
One might try to be charitable and "correct" the question into asking for two points where the probability density is the same -- but that wouldn't lead to $a=60,b=65$ being a solution; instead we would have $a=60+t,b=60-t$ for some t (since the distribution is symmetric around $\mu =60$).
My tentative conclusion would be that (a) it's a trick question, (b) your understanding is correct, and (c) the solution you quote is just meant to be one possible answer.
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Evelyn Freeman
the question is pretty meaningless