According to a recent study, 1 in every 10 women has been a victim of domestic abuse at some point in

The probability that more than 22 of the women sampled have not been the victim of domestic abuse is P=0.537.

Step-by-step explanation:

This problem can be modeled by a binomial random variable.

The probability p can be calculated as the proportion of women that has not been a victim of domestic abuse at some point in her life:

[tex]p=1-\dfrac{X}{n}=1-\dfrac{1}{10}=1-0.1=0.9[/tex]

The sample size is n=25.

We want to calculate the probability that more than 22 of the women of the sample have not been victim of domestic abuse.

The probability that exactly k women have not been victim of domestic abuse can be calculated as:

[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.9^{k} 0.1^{25-k}\\\\\\[/tex]

Then, the probability that more than 22 of the women sampled have not been the victim of domestic abuse is:

[tex]P(x22)=P(x=23)+P(x=24)+P(x=25)\\\\\\P(x=23) = \dbinom{25}{23} p^{23}(1-p)^{2}=300*0.089*0.01=0.266\\\\\\P(x=24) = \dbinom{25}{24} p^{24}(1-p)^{1}=25*0.08*0.1=0.199\\\\\\P(x=25) = \dbinom{25}{25} p^{25}(1-p)^{0}=1*0.072*1=0.072\\\\\\\\P(x22)=0.266+0.199+0.072=0.537[/tex]

According to the given, 1 in every 10 women has been a victim, therefore the probability of a  victim of domestic abuse at some pint is

[tex]p=\frac{1}{10} = 0.10[/tex]

We need to use Binomial distribution to take a randomly sampled of 25 women and asked each whether she has been a victim.

[tex]f(x)= ^nC_x (p)^x (1-p)^{n-x}[/tex]

Where x is the event that shows a victim of domestic abuse at some point.

We need to find the probability of these victim at least 2 has been victim of abuse, so

[tex]P(x=2) = 1- [P(x=0)+P(x=1)][/tex]

[tex]P(x=2) = 1 - [^{25}C_0(0.10)^0(1-0.10)^{25-0}+^{25}C_1(0.10)^1(1-0.10)^{25-1}][/tex]

[tex]P(x=2) = 1 - [0.07198+0.199][/tex]

[tex]P(x=2) = 0.72902[/tex]

So the probability is 0.729