The mean life of a sample of 60 bulbs was 650 and the standard deviation was 8 h. A second sample of 80 bulbs has a mean life of 660 h and standard deviation 7h. Find the over all standard deviation.

Options

A. 8.97
B. 8.98
C. 8.94
D. None of the above

Answer

To find the overall standard deviation of two samples, we can use the following formula:

\[ \sigma = \sqrt{\frac{n_1 \cdot s_1^2 + n_2 \cdot s_2^2}{n_1 + n_2} + \frac{n_1 \cdot n_2 \cdot (\bar{x}_1 - \bar{x}_2)^2}{(n_1 + n_2)^2}} \]

Where:
- \( n_1 \) = number of bulbs in the first sample = 60
- \( s_1 \) = standard deviation of the first sample = 8
- \( \bar{x}_1 \) = mean of the first sample = 650
- \( n_2 \) = number of bulbs in the second sample = 80
- \( s_2 \) = standard deviation of the second sample = 7
- \( \bar{x}_2 \) = mean of the second sample = 660

### Step 1: Calculate \( n_1 \cdot s_1^2 \)
\[ n_1 \cdot s_1^2 = 60 \cdot (8^2) = 60 \cdot 64 = 3840 \]

### Step 2: Calculate \( n_2 \cdot s_2^2 \)
\[ n_2 \cdot s_2^2 = 80 \cdot (7^2) = 80 \cdot 49 = 3920 \]

### Step 3: Calculate \( n_1 + n_2 \)
\[ n_1 + n_2 = 60 + 80 = 140 \]

### Step 4: Calculate the first part of the formula
\[ \frac{n_1 \cdot s_1^2 + n_2 \cdot s_2^2}{n_1 + n_2} = \frac{3840 + 3920}{140} = \frac{7760}{140} = 55.4286 \]

### Step 5: Calculate \( \bar{x}_1 - \bar{x}_2 \)
\[ \bar{x}_1 - \bar{x}_2 = 650 - 660 = -10 \]

### Step 6: Calculate \( n_1 \cdot n_2 \)
\[ n_1 \cdot n_2 = 60 \cdot 80 = 4800 \]

### Step 7: Calculate \( (n_1 + n_2)^2 \)
\[ (n_1 + n_2)^2 = 140^2 = 19600 \]

### Step 8: Calculate the second part of the formula
\[ \frac{n_1 \cdot n_2 \cdot (\bar{x}_1 - \bar{x}_2)^2}{(n_1 + n_2)^2} = \frac{4800 \cdot (-10)^2}{19600} = \frac{4800 \cdot 100}{19600} = \frac{480000}{19600} = 24.4898 \]

### Step 9: Combine both parts and take the square root
\[ \sigma = \sqrt{55.4286 + 24.4898} = \sqrt{79.9184} \approx 8.94 \]

### Final Answer
The overall standard deviation is approximately \( 8.94 \) hours.

---