JEE Statistics Trick: Wrong Observation Corrected in 1 Minute! 🔥

 

❓ Question

FOR:

The mean and standard deviation of 100 observations are 40 and 5.1, respectively.
By mistake, one observation is taken as 50 instead of 40.

If the correct mean and correct standard deviation are $\mu$ and $\sigma$, respectively, then the value of

$$10(\mu + \sigma)$$

is equal to ?

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🖼️ Question Image

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✍️ Short Solution

This is a standard JEE correction-type statistics question.
The idea is simple:

👉 Use the given (wrong) mean and SD to compute the wrong total and wrong sum of squares.
👉 Correct the single wrong observation.
👉 Recalculate the correct mean and correct standard deviation.

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🔹 Step 1 — Find the wrong total (Σx)

Given:

• Number of observations, $n = 100$
  

• Wrong mean $= 40$
  

So,

$$\Sigma x_{\text{wrong}} = 100 \times 40 = 4000$$

One value was taken as 50 instead of 40.

Correct total:

$$\Sigma x_{\text{correct}} = 4000 - 50 + 40 = 3990$$

Hence,

$$\mu = \frac{3990}{100} = 39.9$$

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🔹 Step 2 — Find the wrong sum of squares (Σx²)

Formula:

$$\sigma^2 = \frac{\Sigma x^2}{n} - \bar{x}^2$$

Given:

$$\sigma = 5.1,\quad \bar{x} = 40$$

So,

$$\Sigma x^2_{\text{wrong}} = n(\sigma^2 + \bar{x}^2)$$
$$= 100(5.1^2 + 40^2) = 100(26.01 + 1600) = 162601$$

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🔹 Step 3 — Correct the sum of squares

Wrong value squared:

$$50^2 = 2500$$

Correct value squared:

$$40^2 = 1600$$

So,

$$\Sigma x^2_{\text{correct}} = 162601 - 2500 + 1600 = 161701$$

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🔹 Step 4 — Find the correct standard deviation

$$\sigma^2 = \frac{161701}{100} - (39.9)^2$$
$$= 1617.01 - 1592.01 = 25$$
$$\Rightarrow \sigma = 5$$

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🔹 Step 5 — Compute $10(\mu + \sigma)$

$$\mu + \sigma = 39.9 + 5 = 44.9$$
$$10(\mu + \sigma) = 10 \times 44.9 = 449$$

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✅ Final Answer

$$\boxed{449}$$