Nomrmal probability distribution A nationalised bank has found that has dialy balance available in its savings accounts follows a normal distribution with a mean of rs. 500 and standard deviation of rs. 50 The percentage of saving account holders, who maintain an average dialy balance more thamln rs. 500 is......
Here. P(X>500)=?
According to origin transformation.
P(X>(500-500)/50)
Where E(x)=500 and S.D =50
P(X>0)= 
1- P(-1

=1-0.3453=65.47%
But i got an incorrect answer 
Correct answer is 49.6 %

If we assume perfect normality, then we want
$$\Pr(Z\gt (500-500)/50),$$
that is, $\Pr(Z\gt 0)$, where $Z$ is standard normal. This is exactly $0.5$, since $\Pr(Z=0)=0$. The problem said account amounts are normally distributed, so $0.5$ is the technically correct answer to the problem as stated.
But the proposed answer of $49.6\%$ indicates that they are assuming that bank accounts have an integer number of Rs. Thus they are assuming that the distribution is not really fully normal, and are using a continuity correction. The account has $\gt 500$ Rs., so $\ge 501$. The probability that $X\ge 501$ is, roughly, $1$ minus the probability that the approximating normal is $\le 500.5$. So for our approximation we use
$$1-\Pr(Z\le (500.5-500)/50).$$
We have $(500.5-500)/50=0.01$, and now the standard normal table yields $1-0.5040$.
