<u>Answer:</u>
The probability that at most 10 of them will be living with both of their parents is 0.0018
<u>Solution:</u>
Given, US 1995, 70% of all children in the U.S. were living with both parents. 25 children were selected at random in the U.S.,
We have to find what is the probability that at most 10 of them will be living with both of their parents?
Now, number of children picked (n) = 25.
Probability to be with parents = 70% = 0.7.
Our condition is 0 ≤ x ≤ 10.
Then, we have to calculate p(0 ≤ x ≤ 10).
We can calculate using binomial cumulative distribution frequency- binomial CDF
Now, p(0 ≤ x ≤ 10) = binomcdf(n, p, x) = binomcdf(25, 0.7, 10) = 0.00177840 ≈ 0.0018.
Hence, the probability is 0.0018
