Respuesta :
Answer:
The probability that 100 or fewer will have 3 ​girls is 0.00734.
Step-by-step explanation:
The complete question is:
In a family with ​3 children, the probability that all the children are girls is approximately 0.125. In a random sample of 1000 families with ​3 children, what is the approximate probability that 100 or fewer will have 3 ​girls? Approximate a binomial distribution with a normal distribution.
Solution:
Let X represent the number of families who has 3 girls.
The random variable X follows a Binomial distribution with parameters n = 1000 and p = 0.125.
But the sample selected is too large.
So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
1. np ≥ 10
2. n(1 - p) ≥ 10
Check the conditions as follows:
 [tex]np=1000\times 0.125=125>10\\\\n(1-p)=1000\times (1-0.125)=875>10[/tex]
Thus, a Normal approximation to binomial can be applied.
So, Â [tex]X\sim N(\mu=np,\sigma^{2}=np(1-p))[/tex]
The mean and standard deviation are:
[tex]\mu=np=1000\times 0.125=125\\\\\sigma=\sqrt{np(1-p)}=\sqrt{1000\times 0.125\times (1-0.125)}=10.46[/tex]
Compute the probability that 100 or fewer will have 3 ​girls as follows:
Apply Continuity correction:
[tex]P(X\leq 100)=P(X<100-0.50)[/tex]
         [tex]=P(X<99.50)\\\\=P(\frac{X-\mu}{\sigma}<\frac{99.5-125}{10.46})\\\\=P(Z<-2.44)\\\\=0.00734[/tex]
*Use a z-table.
Thus, the probability that 100 or fewer will have 3 ​girls is 0.00734.
