Question

below are the probabilities associated with x = number of cancer survivors who are centenarians.

x	p(x)
0	0.30
1	0.35

|2|

3	0.15
4 or more	0.0

the cumulative probability of 2 is choose
the probability of no cancer survivors who are centenarians
the probability of 3 or more centenarian cancer survivors
the probability of at most 1 centenarian cancer survivor is
p(x < 2) is
the expected number of centenarian cancer survivors is
if the standard deviation is 1.023, then the variance is
cdf(2) is

Explanation:

Step1: Define cumulative probability formula

The cumulative - probability $P(X\leq k)=\sum_{i = 0}^{k}p(i)$.

Step2: Calculate cumulative probability of 2

$P(X\leq2)=p(0)+p(1)+p(2)$. We know $p(0) = 0.30$, $p(1)=0.35$, and since the sum of all probabilities must be 1, $p(2)=1-(0.30 + 0.35+0.15+0.0)=0.20$. So $P(X\leq2)=0.30 + 0.35+0.20=0.85$.

Step3: Probability of no cancer survivors

The probability of no cancer survivors ($x = 0$) is $p(0)=0.30$.

Step4: Probability of 3 or more

$P(X\geq3)=p(3)+p(4\text{ or more})=0.15 + 0.0=0.15$.

Step5: Probability of at most 1

$P(X\leq1)=p(0)+p(1)=0.30 + 0.35=0.65$.

Step6: Probability $P(X\lt2)$

$P(X\lt2)=p(0)+p(1)=0.30 + 0.35=0.65$.

Step7: Calculate expected value

The expected value $E(X)=\sum_{i}x_ip(x_i)=0\times0.30+1\times0.35 + 2\times0.20+3\times0.15+4\times0+...=0.35 + 0.40+0.45=1.2$.

Step8: Calculate variance from standard - deviation

The variance $\sigma^{2}$ and standard - deviation $\sigma$ are related by $\sigma^{2}=\sigma^{2}$. Given $\sigma = 1.023$, then $\sigma^{2}=(1.023)^{2}=1.046529\approx1.05$.

Answer:

1. The cumulative probability of 2 is 0.85.
2. The probability of no cancer survivors who are centenarians is 0.30.
3. The probability of 3 or more centenarian cancer survivors is 0.15.
4. The probability of at most 1 centenarian cancer survivor is 0.65.
5. $P(X\lt2)$ is 0.65.
6. The expected number of centenarian cancer survivors is 1.2.
7. If the standard deviation is 1.023, then the variance is approximately 1.05.
8. CDF(2) is 0.85.