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For a binomial variate x, it is known that n = 5, p (x = 1) = 0.4096,p (x = 2) = 0.2048. Then the probability distribution function of x is (a) p(x) = 5 Cx (0.8)x (0.2) x – 5 (b) p(x) = 5 Cx(0.2)x (0.8)5 (c) p(x) = 5 Cx (0.2)x (0.8)x – 5 (d) p(x) = 5 Cx (0.2)x (0.8)5 – x?
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For a binomial variate x, it is known that n = 5, p (x = 1) = 0.4096,p...
Solution:
Given, n = 5, p(x = 1) = 0.4096, p(x = 2) = 0.2048.
We know that the probability mass function of the binomial distribution is given by:
P(x) = nCx * p^x * (1-p)^(n-x)
where, x = 0, 1, 2, ..., n.
To find the probability distribution function of x, we need to find the values of p(x=0), p(x=3), p(x=4) and p(x=5) as well.
Let's find the value of p(x=0) first.
We know that the sum of probabilities of all possible values of x is equal to 1.
Therefore,
p(x=0) + p(x=1) + p(x=2) + p(x=3) + p(x=4) + p(x=5) = 1
p(x=0) + 0.4096 + 0.2048 + p(x=3) + p(x=4) + p(x=5) = 1
p(x=0) + p(x=3) + p(x=4) + p(x=5) = 1 - 0.4096 - 0.2048
p(x=0) + p(x=3) + p(x=4) + p(x=5) = 0.3856
Now, we can use the given probabilities of p(x=1) and p(x=2) to find the values of p(x=3), p(x=4) and p(x=5).
We know that,
p(x=1) = 5C1 * p^1 * (1-p)^(5-1)
0.4096 = 5 * p * (1-p)^4
p = 0.2
Similarly,
p(x=2) = 5C2 * p^2 * (1-p)^(5-2)
0.2048 = 10 * p^2 * (1-p)^3
p = 0.8
Now, we can substitute the values of p in the probability mass function to get the probability distribution function of x.
Hence, the correct option is:
p(x) = 5 Cx (0.2)x (0.8)5-x
Explanation:
- Finding p(x=0) using the sum of probabilities formula.
- Using p(x=1) and p(x=2) to find the values of p(x=3), p(x=4) and p(x=5).
- Substituting the values of p in the probability mass function to get the probability distribution function of x.
- Option (d) is the correct option.
Given, n = 5, p(x = 1) = 0.4096, p(x = 2) = 0.2048.
We know that the probability mass function of the binomial distribution is given by:
P(x) = nCx * p^x * (1-p)^(n-x)
where, x = 0, 1, 2, ..., n.
To find the probability distribution function of x, we need to find the values of p(x=0), p(x=3), p(x=4) and p(x=5) as well.
Let's find the value of p(x=0) first.
We know that the sum of probabilities of all possible values of x is equal to 1.
Therefore,
p(x=0) + p(x=1) + p(x=2) + p(x=3) + p(x=4) + p(x=5) = 1
p(x=0) + 0.4096 + 0.2048 + p(x=3) + p(x=4) + p(x=5) = 1
p(x=0) + p(x=3) + p(x=4) + p(x=5) = 1 - 0.4096 - 0.2048
p(x=0) + p(x=3) + p(x=4) + p(x=5) = 0.3856
Now, we can use the given probabilities of p(x=1) and p(x=2) to find the values of p(x=3), p(x=4) and p(x=5).
We know that,
p(x=1) = 5C1 * p^1 * (1-p)^(5-1)
0.4096 = 5 * p * (1-p)^4
p = 0.2
Similarly,
p(x=2) = 5C2 * p^2 * (1-p)^(5-2)
0.2048 = 10 * p^2 * (1-p)^3
p = 0.8
Now, we can substitute the values of p in the probability mass function to get the probability distribution function of x.
Hence, the correct option is:
p(x) = 5 Cx (0.2)x (0.8)5-x
Explanation:
- Finding p(x=0) using the sum of probabilities formula.
- Using p(x=1) and p(x=2) to find the values of p(x=3), p(x=4) and p(x=5).
- Substituting the values of p in the probability mass function to get the probability distribution function of x.
- Option (d) is the correct option.
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For a binomial variate x, it is known that n = 5, p (x = 1) = 0.4096,p (x = 2) = 0.2048. Then the probability distribution function of x is (a) p(x) = 5 Cx (0.8)x (0.2) x – 5 (b) p(x) = 5 Cx(0.2)x (0.8)5 (c) p(x) = 5 Cx (0.2)x (0.8)x – 5 (d) p(x) = 5 Cx (0.2)x (0.8)5 – x?
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For a binomial variate x, it is known that n = 5, p (x = 1) = 0.4096,p (x = 2) = 0.2048. Then the probability distribution function of x is (a) p(x) = 5 Cx (0.8)x (0.2) x – 5 (b) p(x) = 5 Cx(0.2)x (0.8)5 (c) p(x) = 5 Cx (0.2)x (0.8)x – 5 (d) p(x) = 5 Cx (0.2)x (0.8)5 – x? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about For a binomial variate x, it is known that n = 5, p (x = 1) = 0.4096,p (x = 2) = 0.2048. Then the probability distribution function of x is (a) p(x) = 5 Cx (0.8)x (0.2) x – 5 (b) p(x) = 5 Cx(0.2)x (0.8)5 (c) p(x) = 5 Cx (0.2)x (0.8)x – 5 (d) p(x) = 5 Cx (0.2)x (0.8)5 – x? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a binomial variate x, it is known that n = 5, p (x = 1) = 0.4096,p (x = 2) = 0.2048. Then the probability distribution function of x is (a) p(x) = 5 Cx (0.8)x (0.2) x – 5 (b) p(x) = 5 Cx(0.2)x (0.8)5 (c) p(x) = 5 Cx (0.2)x (0.8)x – 5 (d) p(x) = 5 Cx (0.2)x (0.8)5 – x?.
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