*1.*

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

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*i) Find mgf of . *

*ii) Evaluate (− > − > ). *

*iii) Find value of such that (| + | ≥ ) = . . *

*iv) Find − *_{ . }* and convert to .*

*2.*

*Let *_{ } < _{ } < _{ } < _{ } < _{ } be the order statistics of five independent observations _{ }, _{ }, _{ }, _{ }*& *_{ } each from distribution with pdf ( ) = ^{-2x}; ≤ < ∞.

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

(ii) Determine ( ≥ . ).

where, _{ } ( ) = sum∑(^{n}_{r} ) [ ( )] ^{ } [ − ( )] ^{ − },

_{ } ( ) = ( !/( − )!( − )!) [ ( )]^{ − } [ − ( )] ^{ − } ( )

3. Let equals weight of a soap. A random sample of size of yielded with weights , , , , , , , , , , &

grams respectively. Now order them increasingly and find Median, _{ } and _{ }. Is

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

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”

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