In a given are of the country, the ages of all the inhabitants are found to lie

Question

In a given are of the country, the ages of all the inhabitants are found to lie in the range of 0 lessthanorequalto A lessthanorequalto 50, where A is the age of the inhabitant. It is further determined that the probability of an inhabitant having an age in the range from 1 to A+dA is proportional to the quantity. [50A - A^2]dA. Obtain the normalized probability distribution for the ages of the inhabitants in the given area of the country. If an inhabitant is chosen randomly, what is the probability that his or her age will be in the range of 10 lessthanorequalto A lessthanorequalto 20? Compute the average age of the inhabitants of the area. If all the inhabitants over the age of 25 in the area were to move out of the area, obtain an expression giving the new normalized probability for the ages of the remaining inhabitants of the area. Compute the new average age of the remaining inhabitants.

Explanation / Answer

Solution

Back-up Theory

1. If X is proportional to Y, then X = kY, where k is a constant, called constant ofproportionality.

2. If a continuous random variable X taking values in the range (a, b, its pdf (probability density function) f(x) must satisfy 2 conditions:

i) 0 f(x) 1 for all x in (a, b) and (ii) integral of f(x) from a to b must be equal to 1.

3. Average or mean of X = E(X) = integral of {x.f(x)} from a to b ; Variance of X = V(X) = E(X2) – {E(X)}2

4. Normalised variable, Z = {X - E(X)}(sq.rt of V(X)}

Now, to work out the solution,

Preparatory work

Let random A represent the age of the inhabitants. Then, we are given f(a) = k.{50a – a2}, a taking values from (0, 50).

Integrating f(a) w.r.t a from 0 to 50 and simplifying, k = 6/125000 and so the pdf of A is:

f(a) = (6/125000){50a – a2}

E(A) = (6/125000) integral of {50a2 – a3} from 0 to 50 = (6/125000)[{(50.503)/3)} – {(504)/4}]

= 6{(50/3) – (50/4)} = 25.

E(A2) = (6/125000) integral of {50a3 – a4} from 0 to 50 = (6/125000)[{(50.504)/4)} – {(505)/5}]

= 6{(2500/4) – (2500/5)} = 750.

So, V(A) = 125

Part (a)

Normalized variable, Z = (A - 25)/55 and its distribution is given by

f(z) = (6/125000)[{50(A - 25)/55} – {(A - 25)/55}2] . ANSWER

Part (b)

Required probability = (6/125000) x integral of {50a – a2} from10 to 20

Part (c)

As already calculated under Preparatory work, average age of inhabitants = 25. ANSWER (

Solution

Back-up Theory

1. If X is proportional to Y, then X = kY, where k is a constant, called constant ofproportionality.

2. If a continuous random variable X taking values in the range (a, b) and its pdf (probability density function) f(x) must satisfy 2 conditions:

i) 0 f(x) 1 for all x in (a, b) and (ii) integral of f(x) from a to b must be equal to 1.

3. Average or mean of X = E(X) = integral of {x.f(x)} from a to b ; Variance of X = V(X) = E(X2) – {E(X)}2

4. Normalised variable, Z = {X - E(X)}(sq.rt of V(X)}

Now, to work out the solution,

Preparatory work

Let random A represent the age of the inhabitants. Then, we are given f(a) = k.{50a – a2}, a taking values from (0, 50).

Integrating f(a) w.r.t a from 0 to 50 and simplifying, k = 6/125000 and so the pdf of A is:

f(a) = (6/125000){50a – a2}

E(A) = (6/125000) integral of {50a2 – a3} from 0 to 50 = (6/125000)[{(50.503)/3)} – {(504)/4}]

= 6{(50/3) – (50/4)} = 25.

E(A2) = (6/125000) integral of {50a3 – a4} from 0 to 50 = (6/125000)[{(50.504)/4)} – {(505)/5}]

= 6{(2500/4) – (2500/5)} = 750.

So, V(A) = 125

Part (a)

Normalized variable, Z = (A - 25)/55 and its distribution is given by

f(z) = (6/125000)[{50(A - 25)/55} – {(A - 25)/55}2] . If need be this can be simplified to get a concise form. ANSWER

Part (b)

Required probability = (6/125000) x integral of {50a – a2} from10 to 20

Part (c)

As already calculated under Preparatory work, average age of inhabitants = 25. ANSWER

Part (d) and (e)

In foregoing working, change the integration range from (0, 50) to (0, 25) and proceed identically.

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