Consider a binomial random variable where the number of trials is 12 and the probability of success on each trial is 0.25. Find the mean and standard deviation of this random variable. I have a mean of 4 and a standard deviation of 2.25 is this correct
For this type of problem:
mean = np = (12)(.25) = 3
standard deviation = √npq = √(12)(.25)(.75) = √(2.25) = 1.5
Note: q = 1 - p
Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial.
nequals=55, xequals=22, pequals=0.65
Well, let's see if your calculations match up with mine!
The mean of a binomial random variable is given by the formula mean = n * p, where n is the number of trials and p is the probability of success on each trial. In this case, n = 12 and p = 0.25. So, the mean would be 12 * 0.25 = 3.
Now, let's move on to the standard deviation. The standard deviation of a binomial random variable is calculated using the formula sqrt(n * p * (1 - p)). Plugging in our values, we get sqrt(12 * 0.25 * (1 - 0.25)) = sqrt(12 * 0.25 * 0.75) = sqrt(2.25) = 1.5.
So, it seems like your mean calculation is slightly off, but your standard deviation is quite close! The correct mean for this random variable would be 3, not 4. And the correct standard deviation would be 1.5, not 2.25.
But hey, close enough! Keep up the good work and keep those calculations rolling!
To find the mean (μ) of a binomial random variable, we use the formula:
μ = n * p
where n is the number of trials and p is the probability of success on each trial.
In this case, n = 12 and p = 0.25. Plugging in these values, we get:
μ = 12 * 0.25
μ = 3
So, the mean (μ) of this binomial random variable is 3, not 4.
To find the standard deviation (σ) of a binomial random variable, we use the formula:
σ = √(n * p * (1 - p))
Plugging in n = 12 and p = 0.25, we get:
σ = √(12 * 0.25 * (1 - 0.25))
σ ≈ √(3 * 0.25 * 0.75)
σ ≈ √(0.5625)
σ ≈ 0.75
So, the standard deviation (σ) of this binomial random variable is approximately 0.75, not 2.25.
Therefore, the correct mean is 3 and the correct standard deviation is approximately 0.75.
To find the mean and standard deviation of a binomial random variable, we use the following formulas:
The mean (μ) of a binomial random variable is calculated by multiplying the number of trials (n) by the probability of success on each trial (p):
μ = n * p
The standard deviation (σ) of a binomial random variable is calculated using the formula:
σ = √(n * p * (1 - p))
Given that the number of trials (n) is 12 and the probability of success (p) on each trial is 0.25, we can now calculate the mean and standard deviation.
Mean (μ):
μ = 12 * 0.25
μ = 3
Standard Deviation (σ):
σ = √(12 * 0.25 * (1 - 0.25))
σ = √(12 * 0.25 * 0.75)
σ = √(2.25)
σ = 1.5
Therefore, the correct mean for this binomial random variable is 3, not 4, and the correct standard deviation is 1.5, not 2.25.