Answer to Question #245064 in Statistics and Probability for mechanix

Question

Consider the pdf of a Normal variable 𝑿 is defined as 𝒇(𝒙) = (1/√98𝝅)^{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)

The moment generating function (MGF):

$m_X(t)=Ee^{tX}$

$m_X(t)=\int^{\infin}_{-\infin}e^{tx}f(x)dx=\int^{\infin}_{-\infin}\frac{e^{tx}e^{-(x+11)^2/98}}{\sqrt{98\pi}}dx=$

$=\frac{e^{-121/98}}{\sqrt{98\pi}}\int^{\infin}_{-\infin}e^{-(x^2+x(98t+22))/98}dx=\frac{e^{-121/98}}{\sqrt{98\pi}}\cdot e^{(98t+22)^2/392}\sqrt{98\pi}=e^{(24.5t^2+11t)}$

ii)

$P(-3>-X>13)=\int^{-3}_{-\infin}f(-x)dx+\int^{\infin}_{13}f(-x)dx=$

$=\int^{-3}_{-\infin}\frac{e^{-(-x+11)^2/98}}{\sqrt{98\pi}}dx+\int^{\infin}_{13}\frac{e^{-(-x+11)^2/98}}{\sqrt{98\pi}}dx=-\frac{(erf(14/\sqrt{98})-1)\sqrt{98\pi}}{2\sqrt{98\pi}}-\frac{(erf(2/\sqrt{98})-1)\sqrt{98\pi}}{2\sqrt{98\pi}}$

where erf is error function.

$erf(14/\sqrt{98})=0.95,\ erf(2/\sqrt{98})=0.22$

$P(-3>-X>13)=0.025+0.390=0.415$

iii)

$P(|X+11|\ge C)=2\int^{\infin}_{C-11}\frac{e^{-(x+11)^2/98}}{\sqrt{98\pi}}dx=0.0614$

$2\cdot-\frac{erf((\sqrt{2}(C-11)+11\sqrt{2})/14)-1}{2}=0.0614$

$erf((\sqrt{2}(C-11)+11\sqrt{2})/14)=1-0.0614=0.9386$

$(\sqrt{2}(C-11)+11\sqrt{2})/14=1.323$

$C=\frac{14\cdot 1.323-11\sqrt{2}}{\sqrt{2}}+11=13.1$

iv)

$Z_{0.0031}=-3.42$

$-Z_{0.0031}=3.42$

$\frac{x-\mu}{\sigma}=3.42$

$\mu=-11,\ \sigma=7$

$X=3.42\sigma+\mu=3.42\cdot7-11=12.94$