The screws produced by a certain machine were checked by examining samples of 7. The following table shows the distribution of 128 samples according to the number of defective items they contained.
No. of defectives in a sample of 7 is :
`{:(,:,0,1,2,3,4,5,6,7),("No of samples" :,:,7,6,19,35,30,23,7,1):}`
N = 128.
Fit a binomial distributtion and find the expected frequencies if the chance of screw being defective is `(1)/(2)`. Also find the mean and variance of the fitted distribution.

Answer

To solve the problem, we will fit a binomial distribution to the given data and calculate the expected frequencies, mean, and variance.

### Step 1: Understand the Problem
We have a sample of 128 observations, where each observation represents the number of defective screws in a sample of 7 screws. The probability of a screw being defective is given as \( p = \frac{1}{2} \).

### Step 2: Set Up the Binomial Distribution
The number of defective screws in a sample of 7 follows a binomial distribution with parameters:
- \( n = 7 \) (number of trials)
- \( p = \frac{1}{2} \) (probability of success, i.e., a screw being defective)

### Step 3: Calculate the Expected Frequencies
The expected frequency for each number of defective screws can be calculated using the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \]
where \( r \) is the number of defective screws (0 to 7).

We will calculate the expected frequencies for \( r = 0, 1, 2, 3, 4, 5, 6, 7 \).

1. **For \( r = 0 \)**:
\[ P(X = 0) = \binom{7}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^7 = 1 \cdot 1 \cdot \frac{1}{128} = \frac{1}{128} \]
Expected frequency = \( 128 \times \frac{1}{128} = 1 \)

2. **For \( r = 1 \)**:
\[ P(X = 1) = \binom{7}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^6 = 7 \cdot \frac{1}{2} \cdot \frac{1}{64} = \frac{7}{128} \]
Expected frequency = \( 128 \times \frac{7}{128} = 7 \)

3. **For \( r = 2 \)**:
\[ P(X = 2) = \binom{7}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^5 = 21 \cdot \frac{1}{4} \cdot \frac{1}{32} = \frac{21}{128} \]
Expected frequency = \( 128 \times \frac{21}{128} = 21 \)

4. **For \( r = 3 \)**:
\[ P(X = 3) = \binom{7}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^4 = 35 \cdot \frac{1}{8} \cdot \frac{1}{16} = \frac{35}{128} \]
Expected frequency = \( 128 \times \frac{35}{128} = 35 \)

5. **For \( r = 4 \)**:
\[ P(X = 4) = \binom{7}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^3 = 35 \cdot \frac{1}{16} \cdot \frac{1}{8} = \frac{35}{128} \]
Expected frequency = \( 128 \times \frac{35}{128} = 35 \)

6. **For \( r = 5 \)**:
\[ P(X = 5) = \binom{7}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^2 = 21 \cdot \frac{1}{32} \cdot \frac{1}{4} = \frac{21}{128} \]
Expected frequency = \( 128 \times \frac{21}{128} = 21 \)

7. **For \( r = 6 \)**:
\[ P(X = 6) = \binom{7}{6} \left(\frac{1}{2}\right)^6 \left(\frac{1}{2}\right)^1 = 7 \cdot \frac{1}{64} \cdot \frac{1}{2} = \frac{7}{128} \]
Expected frequency = \( 128 \times \frac{7}{128} = 7 \)

8. **For \( r = 7 \)**:
\[ P(X = 7) = \binom{7}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^0 = 1 \cdot \frac{1}{128} \cdot 1 = \frac{1}{128} \]
Expected frequency = \( 128 \times \frac{1}{128} = 1 \)

### Step 4: Summarize Expected Frequencies
The expected frequencies for each number of defectives are:
- \( r = 0 \): 1
- \( r = 1 \): 7
- \( r = 2 \): 21
- \( r = 3 \): 35
- \( r = 4 \): 35
- \( r = 5 \): 21
- \( r = 6 \): 7
- \( r = 7 \): 1

### Step 5: Calculate the Mean and Variance
The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are given by:
- Mean: \( \mu = n \cdot p = 7 \cdot \frac{1}{2} = 3.5 \)
- Variance: \( \sigma^2 = n \cdot p \cdot (1-p) = 7 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{7}{4} = 1.75 \)

### Final Answers
- Expected Frequencies: {1, 7, 21, 35, 35, 21, 7, 1}
- Mean: \( 3.5 \)
- Variance: \( 1.75 \)