Mean and standard deviation of a binomial distribution is 5 and 2 res...
Given that, mean = np = 5
Standard deviation,σ=2 , then variance =
σ2 = npq = 22 = 4
Thus,
q=4/5 and p=1−q=1−4/5=1/5
Mode is an integer such that, np+p>x>np−q
⇒ 5 + 1/5 > x > 5 − 4/5
⇒ 26/5 > x > 21/5
⇒ 5.2 > x > 4.2
⇒ x = 5.
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Mean and standard deviation of a binomial distribution is 5 and 2 res...
Mean and Standard Deviation of Binomial Distribution
A binomial distribution is a probability distribution that describes the number of successes in a fixed number of independent trials, each with the same probability of success. The mean and standard deviation of a binomial distribution can be calculated using the following formulas:
Mean = n * p
Standard Deviation = sqrt(n * p * (1-p))
where n is the number of trials and p is the probability of success in each trial.
Mode of Binomial Distribution
The mode of a probability distribution is the value that occurs most frequently. In a binomial distribution, the mode can be calculated as follows:
Mode = floor((n+1) * p)
where floor() is the floor function, which rounds down to the nearest integer.
Answer
Given that the mean and standard deviation of the binomial distribution are 5 and 2 respectively, we can solve for the values of n and p as follows:
Mean = n * p
5 = n * p
Standard Deviation = sqrt(n * p * (1-p))
2 = sqrt(n * p * (1-p))
4 = n * p * (1-p)
Using the first equation, we can solve for p as:
p = 5 / n
Substituting this into the third equation, we get:
4 = n * (5 / n) * (1 - (5/n))
4 = 5 - 25/n + 5/n
25/n = 1
n = 25
Substituting this back into the equation for p, we get:
p = 5/25 = 0.2
Therefore, the binomial distribution has n = 25 trials with a probability of success of p = 0.2.
Using the formula for the mode of a binomial distribution, we get:
Mode = floor((n+1) * p)
Mode = floor((25+1) * 0.2)
Mode = floor(5.2)
Mode = 5
Therefore, the mode of the binomial distribution is 5.