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a. What is the mean survival time of a randomly selected animal of the type used in the experiment?
b. What is the standard deviation of survival time?
c. What is the probability that an animal survives more than 30 weeks?

1 Answer

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Answer:

Step-by-step explanation:

From the given information:

Let X be the random variable;

The mean survival time is E(X):

E(X) = μ

E(X) = ∝β

E(X) = 5 × 10

E(X) = 50

To find the standard deviation, we must first determine the variance:

V(X) = \sigma ^2

V(X) = \alpha \beta ^2

V(X) = 5 \times (10)^2

V(X) = 5 \times 100

V(X) = 500

The standard deviation is:

\sigma =\sqrt{V(X)}

\sigma =\sqrt{500}

\sigma =22.361

The standard deviation of survival time = 22.361

(c)

P(X>30) = 1 - P(X ≤ 30)

= 1 - \int \limits ^{30}_{0} \ f(x) \ dx

= 1 - \int \limits ^{30}_{0}  \dfrac{1}{\beta^{\alpha}  \Gamma \alpha }x^{a-1} e ^{-x/\beta } \ dx

= 1 - \int \limits ^{30}_{0}  \dfrac{1}{\beta^{\alpha}  \Gamma (5)}x^{5-1} e ^{-x/10 } \ dx  --- (1)

So, if:

y = \dfrac{x}{10}

dy = \dfrac{dx}{10}

10dy = dx

and limits of y are 0 and 3.

because;

for x = 0

y = \dfrac{0}{\beta}

y = 0

For x = 30

y = \dfrac{30}{10}

y = 10

From (1);

P(X> 30) = 1 - \int \limits ^{30}_{0}  \dfrac{1}{\beta^{\alpha}  \Gamma (5)}x^{5-1} e ^{-x/10 } \ dx

P(X> 30) = 1 - \int \limits ^{3}_{0}  \dfrac{(10y)^4}{10^{5}  \Gamma (5)}e ^{-y } \ dy   since y = x/β

P(X> 30) = 1 - \int \limits ^{3}_{0}  \dfrac{10^4 y^4 \ e^{-y} }{10^{4}  \Gamma (5)} \ dy

P(X> 30) = 1 - \int \limits ^{3}_{0}  \dfrac{y^4 \ e^{-y} }{ \Gamma (5)} \ dy

Since;

F(a, \alpha )= \int \limits ^a_0 \dfrac{y^{a-1}e^{-y}}{\Gamma (\alpha )} dy

Then;

P(X> 30) = 1 - F(3,5)

From the table of incomplete gamma function;

P(X > 30) = 1 - 0.185

P(X > 30) = 0.815

Thus, the probability that an animal survives more than 30 weeks is 0.815

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Fanny Morar
15.5k 3 10 26
answered 6 months ago