Question

34% of working mothers do not have enough money to cover their health insurance deductibles. you randomly select six working mothers and ask them whether they have enough money to cover their health insurance deductibles. the random variable represents the number of working mothers who do not have enough money to cover their health insurance deductibles. complete parts (a) through (c) below. (a) construct a binomial distribution using n = 6 and p = 0.34.

x

p(x)

|0|
|1|
|2|
|3|
|4|
|5|
|6|
(round to the nearest thousandth as needed.)

Explanation:

Step1: Recall binomial - probability formula

The binomial - probability formula is $P(x)=C(n,x)\times p^{x}\times(1 - p)^{n - x}$, where $C(n,x)=\frac{n!}{x!(n - x)!}$, $n$ is the number of trials, $x$ is the number of successes, and $p$ is the probability of success on a single trial. Here, $n = 6$ and $p=0.34$, so $1 - p = 0.66$.

Step2: Calculate $P(0)$

$C(6,0)=\frac{6!}{0!(6 - 0)!}=1$, $P(0)=C(6,0)\times(0.34)^{0}\times(0.66)^{6}=1\times1\times0.0828524576\approx0.106$.

Step3: Calculate $P(1)$

$C(6,1)=\frac{6!}{1!(6 - 1)!}=\frac{6!}{1!5!}=6$, $P(1)=C(6,1)\times(0.34)^{1}\times(0.66)^{5}=6\times0.34\times0.1255340267\approx0.294$.

Step4: Calculate $P(2)$

$C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15$, $P(2)=C(6,2)\times(0.34)^{2}\times(0.66)^{4}=15\times0.1156\times0.189= 0.329$.

Step5: Calculate $P(3)$

$C(6,3)=\frac{6!}{3!(6 - 3)!}=20$, $P(3)=C(6,3)\times(0.34)^{3}\times(0.66)^{3}=20\times0.039304\times0.287496\approx0.198$.

Step6: Calculate $P(4)$

$C(6,4)=\frac{6!}{4!(6 - 4)!}=15$, $P(4)=C(6,4)\times(0.34)^{4}\times(0.66)^{2}=15\times0.01336336\times0.4356\approx0.069$.

Step7: Calculate $P(5)$

$C(6,5)=\frac{6!}{5!(6 - 5)!}=6$, $P(5)=C(6,5)\times(0.34)^{5}\times(0.66)^{1}=6\times0.0045435424\times0.66\approx0.013$.

Step8: Calculate $P(6)$

$C(6,6)=\frac{6!}{6!(6 - 6)!}=1$, $P(6)=C(6,6)\times(0.34)^{6}\times(0.66)^{0}=1\times0.00154480449\times1\approx0.001$.

Answer:

x	P(x)
1	0.294
2	0.329
3	0.198
4	0.069
5	0.013
6	0.001