Let \(F(Z)\) be the cumulative density function of the standard normal variate \(Z\), then which of the following are correct?
(A) \(F(Z) = \displaystyle\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{Z}\displaystyle e^{-\frac{z^2}{2}}\,dz,\ -\infty \lt Z \lt \infty\)
(B) \(F(-Z) = 1 - F(Z)\)
(C) \(F(0) = 0\)
(D) \(F(\infty) = 1\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
The standard normal density is \(\phi(z) = \frac{1}{\sqrt{2\pi}}e^{-z^2/2}\), and the cumulative distribution function is \(F(Z) = P(\text{variate} \le Z)\). The curve is symmetric about 0 and the total area under it is 1.
Checking the statements
(A) \(F(Z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{Z} e^{-\frac{z^2}{2}}\,dz,\ -\infty \lt Z \lt \infty\) — Correct. F(Z) is the area under the density from \(-\infty\) up to Z; z is only the variable of integration.
(B) \(F(-Z) = 1 - F(Z)\) — Correct. By symmetry, the area to the left of \(-Z\) equals the area to the right of Z, which is \(1 - F(Z)\).
(C) \(F(0) = 0\) — Incorrect. By symmetry, half of the area lies to the left of 0, so \(F(0) = \frac{1}{2}\).
(D) \(F(\infty) = 1\) — Correct. Up to \(\infty\) the whole area is covered, and it equals 1.