In a biomedical research study, it was determined that the survival time, in weeks, of an animal subjected to a certain exposure of gamma radiation has a gamma distribution with α= 5 and β=10.

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?

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:

[tex]V(X) = \sigma ^2[/tex]

[tex]V(X) = \alpha \beta ^2[/tex]

[tex]V(X) = 5 \times (10)^2[/tex]

[tex]V(X) = 5 \times 100[/tex]

[tex]V(X) = 500[/tex]

The standard deviation is:

[tex]\sigma =\sqrt{V(X)}[/tex]

[tex]\sigma =\sqrt{500}[/tex]

[tex]\sigma =22.361[/tex]

The standard deviation of survival time = 22.361

(c)

P(X>30) = 1 - P(X ≤ 30)

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

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

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

So, if:

[tex]y = \dfrac{x}{10}[/tex]

[tex]dy = \dfrac{dx}{10}[/tex]

10dy = dx

and limits of y are 0 and 3.

because;

for x = 0

[tex]y = \dfrac{0}{\beta}[/tex]

y = 0

For x = 30

[tex]y = \dfrac{30}{10}[/tex]

y = 10

From (1);

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

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

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

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

Since;

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

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

a)The mean survival time of a randomly selected animal of the type used in the experiment is 50.

b)Standard deviation of survival time is 22.36

The standard deviation of a random variable, sample, statistical population, data set, or probability distribution is the square root of its variance.

Suppose x is the random variable.

It is given that α= 5 and β=10.

So, the mean survival E(x)=α β

The mean survival E(x) =5*10

The mean survival E(x) =50

We know that Variance =α β²

So, Variance = 5*10²

Variance =500

We know that standard deviation σ = √Variance

σ = √500

σ = 22.36

Therefore, a)the mean survival time of a randomly selected animal of the type used in the experiment is 50.

b)Standard deviation of survival time is 22.36

To get more about standard deviation visit:

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The standard deviation is:

The standard deviation of survival time = 22.361

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