If for a binomial distribution, the mean is 10 and the variance is 5, then find the parameters and `P(X gt 6)`.

Answer

To solve the problem step by step, we will first identify the parameters of the binomial distribution and then calculate \( P(X > 6) \).

### Step 1: Identify the parameters of the binomial distribution

The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p \]
where \( n \) is the number of trials and \( p \) is the probability of success.

The variance \( \sigma^2 \) of a binomial distribution is given by:
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \) is the probability of failure.

From the problem, we know:
\[ \mu = 10 \quad \text{and} \quad \sigma^2 = 5 \]

### Step 2: Set up the equations

From the mean:
\[ n \cdot p = 10 \quad (1) \]

From the variance:
\[ n \cdot p \cdot q = 5 \quad (2) \]

### Step 3: Substitute \( q \) in terms of \( p \)

Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ n \cdot p \cdot (1 - p) = 5 \]

### Step 4: Substitute \( n \) from equation (1) into equation (2)

From equation (1), we can express \( n \) as:
\[ n = \frac{10}{p} \]

Substituting this into equation (2):
\[ \frac{10}{p} \cdot p \cdot (1 - p) = 5 \]
This simplifies to:
\[ 10(1 - p) = 5 \]
\[ 10 - 10p = 5 \]
\[ 10p = 5 \]
\[ p = \frac{1}{2} \]

### Step 5: Find \( q \)

Now that we have \( p \):
\[ q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 6: Find \( n \)

Now substitute \( p \) back into equation (1) to find \( n \):
\[ n \cdot \frac{1}{2} = 10 \]
\[ n = 20 \]

### Step 7: Summary of parameters

We have found:
\[ n = 20, \quad p = \frac{1}{2}, \quad q = \frac{1}{2} \]

### Step 8: Calculate \( P(X > 6) \)

To find \( P(X > 6) \), we can express it as:
\[ P(X > 6) = P(X = 7) + P(X = 8) + \ldots + P(X = 20) \]

Using the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n - r} \]

Thus:
\[ P(X > 6) = \sum_{r=7}^{20} \binom{20}{r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{20 - r} \]
\[ = \sum_{r=7}^{20} \binom{20}{r} \left(\frac{1}{2}\right)^{20} \]

### Step 9: Factor out the constant

Since \( \left(\frac{1}{2}\right)^{20} \) is constant for all terms, we can factor it out:
\[ P(X > 6) = \left(\frac{1}{2}\right)^{20} \sum_{r=7}^{20} \binom{20}{r} \]

### Step 10: Calculate the sum of combinations

The sum \( \sum_{r=7}^{20} \binom{20}{r} \) can be calculated using the binomial theorem:
\[ \sum_{r=0}^{20} \binom{20}{r} = 2^{20} \]
Thus,
\[ \sum_{r=0}^{6} \binom{20}{r} = 2^{20} - \sum_{r=7}^{20} \binom{20}{r} \]
So,
\[ \sum_{r=7}^{20} \binom{20}{r} = 2^{20} - \sum_{r=0}^{6} \binom{20}{r} \]

### Final Calculation

Now, we can compute \( P(X > 6) \):
\[ P(X > 6) = \left(\frac{1}{2}\right)^{20} \left(2^{20} - \sum_{r=0}^{6} \binom{20}{r}\right) \]

This gives us the final answer for \( P(X > 6) \).