A normal law with mean 2

Topic: Functions of random variables and limit theorems · more questions on this topic
‹ An affine image of a uniform variable State the probability functions ›

Let Y\sim\operatorname{N}(2,5).

  1. Describe the density and the distribution function: centre, spread, and the limits at \pm\infty.

  2. Calculate \mathbb{P}(Y\le 5). Use \Phi(1.34)\approx 0.910.

  3. Let Z=3Y-4. Calculate \mathbb{P}(1\le Z\le 5). Use \Phi(0.15)\approx 0.560 and \Phi(0.45)\approx 0.674.

  1. The density is symmetric about 2, positive on the whole line, and highest at 2. The standard deviation is \sqrt{5}\approx 2.24, so the inflection points sit near 2\pm 2.24. The distribution function increases from 0 to 1, with its steepest slope at 2.

  2. \mathbb{P}(Y\le 5)=\Phi((5-2)/\sqrt{5})=\Phi(3/\sqrt{5})\approx\Phi(1.34)\approx 0.910.

  3. Z\sim\operatorname{N}(2, 45). Standardising the endpoints gives

\mathbb{P}(1\le Z\le 5)=\Phi(3/\sqrt{45})-\bigl(1-\Phi(1/\sqrt{45})\bigr)\approx 0.674-(1-0.560)=0.234.