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The problem states that the average of 174 lab readings was initially recorded as 70 units. Later, it was found that a reading of 244 was wrongly typed as 70. We are to find the correct average.

### Step-by-Step Solution:

1. **Find the total sum of the readings with the incorrect value:**
   The formula for the average is:

   \[
   \text{Average} = \frac{\text{Total Sum of Readings}}{\text{Number of Readings}}
   \]

   From the question, we know:
   \[
   70 = \frac{\text{Total Sum}}{174}
   \]

   Therefore, the total sum of the readings with the incorrect value is:
   \[
   \text{Total Sum} = 70 \times 174 = 12180
   \]

2. **Adjust the total sum to correct the wrong reading:**
   A reading of 244 was incorrectly recorded as 70. To correct this, we subtract the wrong value (70) and add the correct value (244):

   \[
   \text{Corrected Total Sum} = 12180 - 70 + 244 = 12354
   \]

3. **Calculate the new average:**
   Now, using the corrected total sum, the correct average is:

   \[
   \text{New Average} = \frac{12354}{174} \approx 71
   \]

Thus, the correct average is **71**, which matches option (D).

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Would you like further details on any step? Here are some related questions to deepen your understanding:

1. How would the average change if another reading was found incorrect?
2. What is the importance of correcting outliers or incorrect values in datasets?
3. How does the number of data points affect the sensitivity of the average?
4. Can you calculate the new average if two readings are found incorrect?
5. How does removing or adding a single data point impact the mean?

**Tip**: Always verify data points when computing averages to avoid skewed results!

The average of 174 lab readings was 70 units. Later on, it was found that a reading 244 was wrongly typed as 70. Find the correct average.

Learn how to correct the average of 174 lab readings after finding an error in one reading. This problem explains how to recalculate the average by adjusting the total sum, using the arithmetic mean formula. Suitable for Grades 9-10.

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