Question

An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X 5 the number of months between successive payments. The cdf of X is as follows:

0 x<1
0.30 1< x <3
F(x)= 0.40 3< x <4
0.45 4< x <6
0.60 6< x <12
1 12< x

Required:
a. What is the pmf of x?
b. Using just the cdf, compute P(3< x <6)and P(4< x)

Answer 1

Answer:

(a)

$\begin{array}{cccccc}x & {1} & {3} & {4} & {6} & {12} \ \\ P(x) & {0.30} & {0.10} & {0.05} & {0.15} & {0.40} \ \end{array}$

(b)

$P(3 \le x \le 6) = 0.30$

$P(4 \le x)=0.60$

Step-by-step explanation:

Given

$F(x) = \left[\begin{array}{ccc}0& x<1 &\\0.30&1 \le x<3 &\\0.40&3 \le x < 4& &0.45 &4 \le x<6 &\\0.60 & 6 \le x < 12 & & 1 & 12 \le x\end{array}\right$

Solving (a): The pmf

This means that we list out the probability of each value of x.

To do this, we simply subtract the current probability value from the next.

So, we have:

$\begin{array}{cccccc}x & {1} & {3} & {4} & {6} & {12} \ \\ P(x) & {0.30} & {0.10} & {0.05} & {0.15} & {0.40} \ \end{array}$

The calculation is as follows:

$0.30 - 0 = 0.30$

$0.40 - 0.30 = 0.10$

$0.45 - 0.40 = 0.05$

$0.60 - 0.45 = 0.15$

$1 - 0.60 = 0.40$

The x values are gotten by considering where the equality sign is in each range.

$1 \le x < 3$ means $x = 1$

$3 \le x < 4$ means $x = 3$

$4 \le x < 6$ means $x=4$

$6 \le x < 12$ means $x = 6$

$12 \le x$ means $x = 12$

Solving (b):

$(i)\ P(3 \le x \le 6)$

This is calculated as:

$P(3 \le x \le 6) = F(6) - F(3-)$

From the given function

$F(6)= 0.60$

$F(3-) = F(1) = 0.30$

So:

$P(3 \le x \le 6) = 0.60 - 0.30$

$P(3 \le x \le 6) = 0.30$

$(ii)\ P(4 \le x)$

This is calculated as:

$P(4 \le x)=1 - F(4-)$

$P(4 \le x)=1 - F(3)$

$P(4 \le x)=1 - 0.40$

$P(4 \le x)=0.60$