Answer to Question #245067 in Statistics and Probability for mechanix

Question

1.

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

i) Find mgf of 𝑿.

ii) Evaluate 𝑷(−𝟑 > −𝑿 > 𝟏𝟑).

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

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

2.

Let 𝒀_𝟏 < 𝒀_𝟐 < 𝒀_𝟑 <𝒀_𝟒 < 𝒀_𝟓 be the order statistics of five independent observations 𝑿_𝟏, 𝑿_𝟐, 𝑿_𝟑, 𝑿_𝟒 & 𝑿_𝟓 each from distribution with pdf 𝒇(𝒙) = 𝒆^-2x; 𝟎 ≤ 𝒙 < ∞.

(i) Find pdf of the sample median and it’s mean.

(ii) Determine 𝑷(𝒀𝟒 ≥ 𝟎. 𝟓). 

where, 𝑮_𝒓 (𝒚) = \sum∑(ⁿ_r ) [𝑭(𝒚)] ^𝒓 [𝟏 − 𝑭(𝒚)] ^𝒏−𝒓 ,

𝒈_𝒓 (𝒚) = (𝒏!/(𝒏−𝒓)!(𝒓−𝟏)!) [𝑭(𝒚)] ^𝒓−𝟏 [𝟏 − 𝑭(𝒚)] ^𝒏−𝒓𝒇(𝒚) 

3. Let 𝑿 equals weight of a soap. A random sample of size 𝟏𝟐 of 𝑿 yielded with weights 𝟓𝟏𝟑, 𝟒𝟗𝟑, 𝟒𝟗𝟖, 𝟓𝟎𝟑, 𝟒𝟗𝟓, 𝟓𝟏𝟏, 𝟓𝟎𝟓, 𝟒𝟗𝟕, 𝟒𝟖𝟖, 𝟒𝟗𝟑, 𝟓𝟏𝟏 & 𝟓𝟎𝟑

grams respectively. Now order them increasingly and find Median, 𝑫_𝟕 and 𝑷_𝟒𝟑. Is

there any mode exists? Determine the semi-range of given weights.

Answer 1

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)

$Z_{0.0031}=-3.42$

$-Z_{0.0031}=3.42$

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

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

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

3.

488, 493, 493, 495, 497, 498, 503, 503, 505, 511, 511, 513

$median=(498+503)/2=500.5$

Decile:

$D_i=i(N+1)/10$

$D_7=7\cdot (12+1)/10=9.1$

Percentile:

$0.43N=0.43\cdot12=5.16$

$0.16\cdot(498-497)=0.16$

$P_{43}=5+0.16=5.16$

Modes are: 493, 503, 511

The semi-interquartile range is one-half the difference between the first and third quartiles:

$IQR=(Q_3-Q_3)/2=(511-493)/2=9$

2.

i)

pdf of the sample median:

$q(x)=c_n[P(x_{med})(1-P(x_{med}))]^{(n-1)/2}f(x)$

$P(x_{med})=0.5$

c_n is a coefficient that represents the number of ways a sample of (n-1)/2 values above median and (n-1)/2 below median can be arranged.

$c_n=2!=2$

$q(x)=2\cdot[0.5\cdot(1-0.5)]^{(5-1)/2}e^{-2x}=0.125e^{-2x}$

The expected value of the median of the sample is equal to the median of the distribution f(x).

$P(x_{med})=\int^{x_{med}}_0f(t)dt=\int^{x_{med}}_0e^{-2t}dt=0.5$

$-e^{-2t}/2|^{x_{med}}_0=-e^{-2x_{med}}/2+1/2=0.5$

$E(X)=x_{med}=\infin$

ii)

$g_4(y)=\frac{5!}{3!}\cdot0.5^3\cdot0.5e^{-2y}=1.25e^{-2y}$

$P(Y_4\ge0.5)=\int^{\infin}_{0.5}g_4(y)dy=1.25\int^{\infin}_{0.5}e^{-2y}dy=$

$=-1.25e^{-2y}/2|^{\infin}_{0.5}=1.25/(2e)=0.2299$