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JEE Mathematics 2025 Question with Solution

The mean and standard deviation of 100100 observations are 4040 and 5.15.1, respectively. By mistake one observation is taken as 5050 instead of 4040. If the correct mean and the correct standard deviation are μ\mu and σ\sigma respectively, then 10(μ+σ)10(\mu + \sigma) is equal to:

  • A

    445445

  • B

    451451

  • C

    447447

  • D

    449449

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The mean of 100100 observations is 4040 and the standard deviation is 5.15.1. One observation was taken as 5050 instead of 4040.

Find: The value of 10(μ+σ)10(\mu + \sigma), where μ\mu and σ\sigma are the correct mean and standard deviation.

First compute the incorrect sum:

∑xi=100×40=4000\sum x_i = 100 \times 40 = 4000

Since 5050 was used instead of 4040, the correct sum is

3990=4000−50+403990 = 4000 - 50 + 40

Therefore, the correct mean is

μ=3990100=39.9\mu = \frac{3990}{100} = 39.9

Now use the variance relation for the incorrect data:

(5.1)2=∑xi2100−(40)2(5.1)^2 = \frac{\sum x_i^2}{100} - (40)^2

So,

∑xi2=100((40)2+(5.1)2)=162601\sum x_i^2 = 100\left((40)^2 + (5.1)^2\right) = 162601

Correct the sum of squares by replacing 5050 with 4040:

∑xi2(correct)=162601−502+402\sum x_i^2\text{(correct)} = 162601 - 50^2 + 40^2 =162601−2500+1600=161701= 162601 - 2500 + 1600 = 161701

Hence,

σ2=161701100−(39.9)2\sigma^2 = \frac{161701}{100} - (39.9)^2 =1617.01−1592.01=25= 1617.01 - 1592.01 = 25

Therefore,

σ=5\sigma = 5

Now,

10(μ+σ)=10(39.9+5)=10(44.9)=44910(\mu + \sigma) = 10(39.9 + 5) = 10(44.9) = 449

Therefore, the correct option is D.

Using corrected mean and variance

Given: Incorrect mean =40= 40, incorrect standard deviation =5.1= 5.1, number of observations =100= 100.

Find: The corrected value of 10(μ+σ)10(\mu + \sigma).

From the mean,

μ=100(40)−50+40100\mu = \frac{100(40) - 50 + 40}{100} μ=40−110=39.9\mu = 40 - \frac{1}{10} = 39.9

The incorrect variance is

(5.1)2=∑xi2100−(xˉ)2(5.1)^2 = \frac{\sum x_i^2}{100} - (\bar{x})^2

Thus,

∑xi2=100×(402)+100×(5.1)2=162601\sum x_i^2 = 100 \times (40^2) + 100 \times (5.1)^2 = 162601

After correction,

σ2=∑xi2−502+402100−(μ)2\sigma^2 = \frac{\sum x_i^2 - 50^2 + 40^2}{100} - (\mu)^2 σ2=1617.01−(39.9)2=25\sigma^2 = 1617.01 - (39.9)^2 = 25

So,

σ=5\sigma = 5

Finally,

10(μ+σ)=10(39.9+5)=44910(\mu + \sigma) = 10(39.9 + 5) = 449

Therefore, the required value is 449449 and the correct option is D.

Common mistakes

  • Using the incorrect sum 40004000 as the correct sum is wrong because one observation was recorded as 5050 instead of 4040. Correct the total first by subtracting 5050 and adding 4040.

  • Correcting the mean but not correcting ∑xi2\sum x_i^2 is wrong because the variance depends on squares of observations. Replace 50250^2 by 40240^2 in the sum of squares.

  • Using the incorrect mean 4040 while computing the corrected variance is wrong because variance must be calculated with the corrected mean μ=39.9\mu = 39.9. Always use the corrected mean after fixing the data.

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