Let X be a binomially distributed variate with mean 10 and variance 5. Then `P(X gt 10)=`

Options

A. `(1)/(2^(20))sum_(11)^(20)20_(C_(K))`
B. `(1)/(2^(20))sum_(1)^(20)-20_(C_(K))`
C. `(1)/(2^(20))sum_(1)^(20)10_(C_(K))`
D. `sum_(11)^(20)20_(C_(K))(1)/(2^(K))((2)/(3))`

Answer

To solve the problem, we need to find \( P(X > 10) \) for a binomially distributed random variable \( X \) with a mean of 10 and a variance of 5.

### Step 1: Understand the parameters of the binomial distribution
The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are given by:
- Mean: \( \mu = Np \)
- Variance: \( \sigma^2 = Np(1-p) \)

Where:
- \( N \) is the number of trials
- \( p \) is the probability of success in each trial

### Step 2: Set up the equations
From the problem, we have:
1. \( Np = 10 \) (mean)
2. \( Np(1-p) = 5 \) (variance)

### Step 3: Substitute \( Np \) into the variance equation
From the first equation, we can express \( Np \) as 10:
\[ Np(1-p) = 5 \implies 10(1-p) = 5 \]

### Step 4: Solve for \( p \)
Now, we can solve for \( p \):
\[ 10 - 10p = 5 \implies 10p = 5 \implies p = \frac{1}{2} \]

### Step 5: Substitute \( p \) back to find \( N \)
Now that we have \( p \), we can substitute it back into the mean equation to find \( N \):
\[ Np = 10 \implies N \cdot \frac{1}{2} = 10 \implies N = 20 \]

### Step 6: Set up the probability expression
Now we know \( N = 20 \) and \( p = \frac{1}{2} \). We need to find \( P(X > 10) \):
\[ P(X > 10) = 1 - P(X \leq 10) \]

### Step 7: Calculate \( P(X \leq 10) \)
Using the binomial probability formula:
\[ P(X = k) = \binom{N}{k} p^k (1-p)^{N-k} \]
we can calculate \( P(X \leq 10) \):
\[ P(X \leq 10) = \sum_{k=0}^{10} \binom{20}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{20-k} = \sum_{k=0}^{10} \binom{20}{k} \left(\frac{1}{2}\right)^{20} \]

### Step 8: Simplify the expression
Thus, we can factor out \( \left(\frac{1}{2}\right)^{20} \):
\[ P(X \leq 10) = \left(\frac{1}{2}\right)^{20} \sum_{k=0}^{10} \binom{20}{k} \]

### Step 9: Use the symmetry of the binomial distribution
Since \( p = \frac{1}{2} \), the binomial distribution is symmetric around its mean, hence:
\[ P(X \leq 10) = P(X \geq 10) = \frac{1}{2} \]

### Step 10: Calculate \( P(X > 10) \)
Now we can find \( P(X > 10) \):
\[ P(X > 10) = 1 - P(X \leq 10) = 1 - \frac{1}{2} = \frac{1}{2} \]

### Final Answer:
\[ P(X > 10) = \frac{1}{2} \]