U1995, 70% of all children in the u. s. were living with both parents. if 25 children were selected at random in the u. s., what is the probability that at most 10 of them will be living with both of their parents? round your answer to 4 decimal places. to answer the question input only the actual number. do not include units. do not give your answer in sentence form -- just include the numerical answer rounded to exactly 4 decimal places.

The probability that at most 10 of them will be living with both of their parents is 0.0018

Solution:

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

25.2

step-by-step explanation:

2+4x6-8/10= 25.2

the answer is 844

step-by-step explanation:

it is 844 because if you add 836.50 + 7.5 you will get 844