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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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