If in Binomial distribution mean is 10 and standard deviation 2, q will be

In a binomial distribution, the mean (μ) and standard deviation (σ) are related to the parameters of the distribution by the following formulas:

μ = n * p
σ = sqrt(n * p * (1 - p))

Where:
- n is the number of trials.
- p is the probability of success on each trial.

Given that the mean (μ) is 10 and the standard deviation (σ) is 2, you can use these formulas to find the values of n and p:

From μ = n * p, you have 10 = n * p.
From σ = sqrt(n * p * (1 - p)), you have 2 = sqrt(n * p * (1 - p)).

You can solve for p in the first equation (10 = n * p) and substitute it into the second equation to find n:

10 = n * p
n = 10 / p

Substituting this into the second equation:

2 = sqrt((10 / p) * p * (1 - p))

Now, you can solve for p:

2 = sqrt(10 * (1 - p))

Square both sides:

4 = 10 * (1 - p)

Now, solve for p:

4/10 = 1 - p
0.4 = 1 - p

p = 1 - 0.4
p = 0.6

So, the probability of success on each trial (p) is 0.6.

Now that you know p, you can find n using the first equation:

10 = n * 0.6
n = 10 / 0.6
n ≈ 16.67

Since n must be a whole number (it represents the number of trials), you should round n to the nearest whole number. In this case, n is approximately 17.

Therefore, in a binomial distribution with a mean of 10 and a standard deviation of 2, n is approximately 17, and p is 0.6.