Question

one with u = 5. An insurance company issues 1250 vision policies. The number of claims filed by a policyholder under a vision policy during year is a random Variable 2 and 6=52. Assume humber of claims filed by different policy holders are mutually independent, Calculate the approximate probability that there is a total of between 2450 and 2600 claims during a one year periode 6. the profit for a new product is given by..Z - 3X-Y-5 Jf Xand Y are independent random variables with x = 1 and by = 2 Find variance of a

Answer 1

5)

Answer:

Probability = 0.0338

Explanation:

Let the random variable X = number of claims by a single policyholder

The random variable X is normally distributed as shown below,

$X\sim N(\mu,\sigma^2)$

$X\sim N(2,(\sqrt{2})^2)$

For n = 1250 vision policies, using the properties of random variables,

$Y (=nX)\sim N(n\mu,n^2\sigma^2)$

$Y (=1250X)\sim N(2500 ,1767.767^2)$

Now, the probability is obtained by converting the normal distribution to the standard normal as shown below,

$P(2450\leq Y\leq 2600)=P(Y\leq 2600)-P(Y\leq 2450)$

$P(2450\leq Y\leq 2600)\Rightarrow P\left (Z\leq \frac{2600-2500}{1767.767} \right )-P\left (Z\leq \frac{2450-2500}{1767.767} \right )$

$P(2450\leq Y\leq 2600)\Rightarrow P\left (Z\leq 0.0566 \right )-P\left (Z\leq -0.0283 \right )$

From the z distribution table,

$P(2450\leq Y\leq 2600)\Rightarrow 0.5226-0.4887=0.0338$

6)

Answer:

The variance of Z = 7

Explanation:

The random variable Z is defined as,

$Z=3X-Y-5$

Using the properties of random variables,

$\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)=\sigma_X^2+\sigma_Y^2$

$\text{Var}(cX+d)=c^2\text{Var}(X)=c^2\sigma_X^2$

now,

$\text{Var}(Z)=\text{Var}(3X-Y-5)=3^2\text{Var}(X)-\text{Var}(Y)=3^2\times (1)^2-(\sqrt{2})^2$

$\text{Var}(Z)=7$