normal distribution - area under the curve

Question

The area (in percentage) under standard normal distribution curve of random variable Z within limits from -3 to 3 is ____.

Please provide some hints on how to solve this problem.

Answer 1

You can look up the probability density for z-values between -3 and +3.In the attached snapshots you can see that the P(0<z<3)=0.4987. Since the standard normal distribution is symmetrical, P(-3<z<0)=0.4987 also. Your question is the probability density in between z-values -3 and +3. In other words, P(-3<z<+3), which is the sum of the two densities I highlighted earlier i.e. P(-3<z<+3)=P(-3<z<0)+P(0<z<3)=0.9974

Density table pic 1

Density table pic 2

Answer 2

Hint: the area satisfying $z\le a$ is $\Phi(a)$, where $\Phi$ is the $N(0,\,1)$ cdf. So what's the area satisfying $a\le z\le b$?

Answer 3

It is $$A={1\over \sqrt {2\pi}}\int_{-3}^3 e^{-x^2\over 2}dx$$which can only be calculated numerically using Q-function as follows $$A=Q(-3)-Q(3)=1-2Q(3)\approx 99.74\%$$