Question

37​% 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. Construct a binomial distribution using n 6 and p = 0.37.
x P(x)
0
1
2
3
4
5
6

Answer 1

Answer:

x         P(x)

0         0.0625

1          0.2203

2         0.3235

3         0.2533

4         0.1116

5         0.0262

6        0.0026

Step-by-step explanation:

If the variable x follows a binomial distribution, the probability P(x) is calculated as:

$P(x)=nCx*p^{x}*(1-p)^{n-x}$

Where $nCx=\frac{n!}{x!(n-x)!}$

Where p is the probability of success (37​% of working mothers do not have enough money to cover their health insurance deductibles), n is the number of identical and independent events (you randomly select six working mothers) and x is the number of that events that has success.

So, replacing n by 6 and p by 0.37, we can calculated the following probabilities as:

$P(x)=6Cx*0.37^{x}*(1-0.37)^{6-x}\\P(0)=6C0*0.37^{0}*(1-0.37)^{6-0}=0.0625\\P(1)=6C1*0.37^{1}*(1-0.37)^{6-1}=0.2203\\P(2)=6C2*0.37^{2}*(1-0.37)^{6-2}=0.3235\\P(3)=6C3*0.37^{3}*(1-0.37)^{6-3}=0.2533\\P(4)=6C4*0.37^{4}*(1-0.37)^{6-4}=0.1116\\P(5)=6C5*0.37^{5}*(1-0.37)^{6-5}=0.0262\\P(6)=6C6*0.37^{6}*(1-0.37)^{6-6}=0.0026$