Question

assume that the proportion of voters who prefer candidate a is $p = 0.207$. organization d conducts a poll of $n = 5$ voters.
let $x$ represent the number of voters polled who prefer candidate a. use some form of appropriate technology (e.g., your calculator or statistics software like excel, r, or statdisk) to find the cumulative probability distribution.
(report answers accurate to 4 decimal places.)

$k$	$p(xleq k)$

|0|
|1|
|2|
|3|
|4|
|5|

Explanation:

Step1: Identify distribution type

This is a binomial distribution problem with $n = 5$ (number of trials) and $p=0.207$. The cumulative - distribution function of a binomial distribution is given by $P(X\leq k)=\sum_{i = 0}^{k}\binom{n}{i}p^{i}(1 - p)^{n - i}$, where $\binom{n}{i}=\frac{n!}{i!(n - i)!}$.

Step2: Calculate $P(X\leq0)$

Using the binomial - probability formula, when $k = 0$, $P(X\leq0)=\binom{5}{0}(0.207)^{0}(1 - 0.207)^{5-0}=(1)\times(1)\times(0.793)^{5}\approx0.3097$.

Step3: Calculate $P(X\leq1)$

$P(X\leq1)=P(X = 0)+P(X = 1)$.
$P(X = 0)=(0.793)^{5}\approx0.3097$, $P(X = 1)=\binom{5}{1}(0.207)^{1}(1 - 0.207)^{4}=5\times0.207\times(0.793)^{4}\approx0.4060$.
So $P(X\leq1)=0.3097 + 0.4060=0.7157$.

Step4: Calculate $P(X\leq2)$

$P(X\leq2)=P(X\leq1)+P(X = 2)$.
$P(X = 2)=\binom{5}{2}(0.207)^{2}(1 - 0.207)^{3}=\frac{5!}{2!(5 - 2)!}\times(0.207)^{2}\times(0.793)^{3}=10\times0.042849\times0.49827\approx0.2133$.
So $P(X\leq2)=0.7157+0.2133 = 0.9290$.

Step5: Calculate $P(X\leq3)$

$P(X\leq3)=P(X\leq2)+P(X = 3)$.
$P(X = 3)=\binom{5}{3}(0.207)^{3}(1 - 0.207)^{2}=10\times0.008869\times0.62885\approx0.0558$.
So $P(X\leq3)=0.9290+0.0558 = 0.9848$.

Step6: Calculate $P(X\leq4)$

$P(X\leq4)=P(X\leq3)+P(X = 4)$.
$P(X = 4)=\binom{5}{4}(0.207)^{4}(1 - 0.207)^{1}=5\times0.001836\times0.793\approx0.0073$.
So $P(X\leq4)=0.9848+0.0073 = 0.9921$.

Step7: Calculate $P(X\leq5)$

$P(X\leq5)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)+P(X = 5)$.
$P(X = 5)=\binom{5}{5}(0.207)^{5}(1 - 0.207)^{0}=(0.207)^{5}\approx0.00079$.
So $P(X\leq5)=0.9921+0.00079 = 1.0000$.

Answer:

$k$	$P(X\leq k)$
$1$	$0.7157$
$2$	$0.9290$
$3$	$0.9848$
$4$	$0.9921$
$5$	$1.0000$