Question

Consider the pdf of a Normal variable 𝑿 is defined as f(x)= (1/√(98pi)^e^-((x+11)^2/98)

i) Find the mgf of 𝑿.

ii) Evaluate 𝑷(−𝟑 > −𝑿 > 𝟏𝟑).

iii) Find the value of 𝑪 such that 𝑷(|𝑿 + 𝟏𝟏| ≥ 𝑪) = 𝟎. 𝟎𝟔𝟏𝟒.

iv) Find −𝒁_{𝟎.𝟎𝟎𝟑𝟏} and convert to 𝑿.

Answer 1

i)

$$N(x; \mu,\sigma^2) =\dfrac{1}{\sqrt{2\pi\sigma^2}} e^{-{1 \over 2}(x-\mu)^2/\sigma^2}$$

Then $\mu=-11, \sigma^2=49.$

The moment generating function corresponding to the normal probability density function$N(x; µ, σ^2)$ is the function $M_{x}(t) = exp\{{\mu t + σ^2t ^2/2}\}$

$$M_{x}(t) = exp\{{-11 t + 49t ^2/2}\}$$

ii)

$$P(-3>-X>13)=P(-13<X<3)$$

$$=P(X<3)-P(X\leq-13)$$

$$=P(Z<\dfrac{3-(-11)}{7})-P(Z\leq\dfrac{-13-(-11)}{7})$$

$$=P(Z<2)-P(Z\leq-0.2857)$$

$$=0.9772499-0.3875485=0.589701$$

iii)

$$P(|X+C|\geq11)=0.0614$$

$$P(X\leq-11-C)=0.0307$$

$$P(Z\leq\dfrac{-11-C-(-11)}{7})=0.0307$$

$$P(Z\leq\dfrac{-C}{7})=0.0307$$

$$\dfrac{-C}{7}=-1.870604$$

$C=13.094228$

iv)

$$P(Z\leq z)=0.0031$$

$$z=-2.737012$$

$$-z=2.737012$$

$$\dfrac{x-(-11)}{7}=-2.737012$$

$$x=-30.159$$

$$-x=30.159$$