Question

24. An insurance company offers its policyholders a num- ber of different premium payment options. For a ran- domly selected policyholder, let $X$ = the number of months between successive payments. The cdf of $X$ is as follows: $F(x) = egin{cases} 0 & x < 1 \ .30 & 1 le x < 3 \ .40 & 3 le x < 4 \ .45 & 4 le x < 6 \ .60 & 6 le x < 12 \ 1 & 12 le x end{cases}$ a. What is the pmf of $X$? b. Using just the cdf, compute $P(3 le X le 6)$ and $P(4 le X)$.

          24. An insurance company offers its policyholders a num-
ber of different premium payment options. For a ran-
domly selected policyholder, let $X$ = the number of
months between successive payments. The cdf of $X$ is as
follows:
$F(x) = egin{cases}
0 & x < 1 \
.30 & 1 le x < 3 \
.40 & 3 le x < 4 \
.45 & 4 le x < 6 \
.60 & 6 le x < 12 \
1 & 12 le x
end{cases}$
a. What is the pmf of $X$?
b. Using just the cdf, compute $P(3 le X le 6)$ and
$P(4 le X)$.

24. An insurance company offers its policyholders a num-
ber of different premium payment options. For a ran-
domly selected policyholder, let X = the number of
months between successive payments. The cdf of X is as
follows:
F(x) = egincases
0 x < 1 .30 1 le x < 3 .40 3 le x < 4 .45 4 le x < 6 .60 6 le x < 12 1 12 le x
endcases
a. What is the pmf of X?
b. Using just the cdf, compute P(3 le X le 6) and
P(4 le X).

Added by Elizabeth K.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Adi S

05/27/2024

Step 1

We can do this by taking the difference between the cdf values at each interval. P(X=0) = F(0) = 0 (since there is no interval for X=0) P(X=1) = F(1) - F(0) = 30% P(X=2) = F(3) - F(1) = 10% P(X=3) = F(4) - F(3) = 5% P(X=4) = F(6) - F(4) = 15% P(X=5) = F(12) - Show more…

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An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows: F(x) = { 0 x < 1 .30 1 <= x < 3 .40 3 <= x < 4 .45 4 <= x < 6 .60 6 <= x < 12 1 12 <= x a. What is the pmf of X? b. Using just the cdf, compute P(3 <= X <= 6) and P(4 <= X).