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Let X1,X2,...,X32 be a random sample from an exponential distribution with a mean of 4. Find an approximate probability that the sample mean is less than 5 (round off to third decimal place).

Sagot :

The probability that the sample mean is less than 5, is 0.92135.

Let X1,X2,...,X32 be a random sample from an exponential distribution with a mean of 4.

We have to find an approximate probability that the sample mean is less than 5.

Let X = [tex]\frac{\Sigma_{i=1}^{n}x_{i}}{n}[/tex]

X = [tex]\frac{\Sigma_{i=1}^{32} x_{i}}{32}[/tex]

Then required probability is P(X<5).

Now we have to find the distribution of X where [tex]x_{i}[/tex]……. x_{32}.

Follow exponential with mean 4.

[tex]x_{i}[/tex]→exponent(λ=1/mean)

[tex]x_{i}[/tex]→exponent(λ=1/4)

[tex]x_{i}[/tex]→exponent(λ=0.25)

Now we have result:

If [tex]x_{1}[/tex] ………… x_{n} are exponential with α.

Then [tex]\Sigma{x_{i}}[/tex] has Gamma Distribution with (α=0.25, n=32)

Now we have result

X→G(α, λ) then c X→G(α/c, λ)

Here, we have

[tex]\frac{\Sigma x_{i}}{n}[/tex]→Gamma(α=0.25, n=32)

[tex]\frac{\Sigma x_{i}}{n}[/tex]→Gamma(α/(1/n), n)

[tex]\frac{\Sigma x_{i}}{n}[/tex]→Gamma(nα, n)

X→Gamma(32*0.25, 32)

X has Gamma distribution with α=8, λ=32.

Required probability is;

P(X<5) = [tex]\int_{0}^{5}f(x) dx[/tex]

P(X<5) =[tex]\int_{0}^{5}\frac{\alpha }{\sqrt{\lambda}}e^{-\alpha x}x^{\lambda-1} dx[/tex]

Now to approximate this probability using normal as

X-mean(X)/SD(X)  = N(0,1)

X→Gamma(α=8, λ=32)

E(X) = λ/α                                            Var(X)= λ/α^2

E(X) = 32/8                                          Var(X)= 32/(8)^2

E(X) = 4                                               Var(X)= 0.5

Now P(X<5)=P[{(X-E(X))/√Var(X)}<{ (5-4)/√0.5}]

P(X<5)=P[{z< 1/√0.5}]

P(X<5) = P(z<1.4142

From the normal probability table

P(X<5) = 0.92135

Hence, the probability that the sample mean is less than 5, is 0.92135.

To learn more about probability link is here

brainly.com/question/11234923

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