Answer:
a)Both X and Y can be well approximated by normal random variables.
Step-by-step explanation:
For each individual, there are only two possible outcomes. Either they are right-handed, or they are left-handed. This means that we can solve this problem using concepts of the binomial probability distribution.
Binomial probability:
Probability of exactly x sucesses on n repeated trials, with probability p.
The binomial probability can be well approximated by normal random variables, using the expected value and the standard deviation
Let X be the number of males (out of the 100) who are left-handed.
and . Can be well approximated.
Let Y be the number of females (out of the 80) who are left-handed.
and . Can be well approximated.
The correct answer is
a)Both X and Y can be well approximated by normal random variables.
Answer with Step-by-step explanation:
We are given that A and B are two countable sets
We have to show that if A and B are countable then is countable.
Countable means finite set or countably infinite.
Case 1: If A and B are two finite sets
Suppose A={1} and B={2}
={1,2}=Finite=Countable
Hence, is countable.
Case 2: If A finite and B is countably infinite
Suppose, A={1,2,3}
B=N={1,2,3,...}
Then, ={1,2,3,....}=N
Hence, is countable.
Case 3:If A is countably infinite and B is finite set.
Suppose , A=Z={..,-2,-1,0,1,2,....}
B={-2,-3}
=Z=Countable
Hence, countable.
Case 4:If A and B are both countably infinite sets.
Suppose A=N and B=Z
Then,==Z
Hence, is countable.
Therefore, if A and B are countable sets, then is also countable.
Answer:
I think its one mean, if it's not I'm very sorry
Another one where there's a danger of overthinking it !
The sum of ANY two integers is odd only if one is even and one is odd.
If they're both odd OR both even, then their sum is even.
So if their sum is odd, then one is odd and the other is even.
The probability is ' 1 ' (100%) .