The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On respectively, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is
- \( \sqrt{3.86} \)
- 1.8
- \( \sqrt{3.96} \)
- 1.94
The Correct Option is C
Solution and Explanation
The given mean is:
\[ \bar{x} = 10 \implies \frac{\Sigma x_i}{20} = 10. \]
Thus:
\[ \Sigma x_i = 10 \times 20 = 200. \]
When the incorrect observation (8) is replaced with the correct value (12):
\[ \Sigma x_i = 200 - 8 + 12 = 204. \]
The corrected mean is:
\[ \bar{x} = \frac{\Sigma x_i}{20} = \frac{204}{20} = 10.2. \]
The standard deviation (S.D.) is given as:
\[ \text{S.D.}^2 = \text{Variance} = 2^2 = 4. \]
From the variance formula:
\[ \frac{\Sigma x_i^2}{20} - \left(\frac{\Sigma x_i}{20}\right)^2 = 4. \]
Substitute:
\[ \frac{\Sigma x_i^2}{20} - 10^2 = 4. \] \[ \frac{\Sigma x_i^2}{20} = 104 \implies \Sigma x_i^2 = 2080. \]
After replacing 8 with 12:
\[ \Sigma x_i^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160. \]
The corrected variance is:
\[ \frac{\Sigma x_i^2}{20} - \left(\frac{\Sigma x_i}{20}\right)^2. \] \[ \frac{2160}{20} - (10.2)^2. \] \[ \frac{\Sigma x_i^2}{20} = 108, \quad (10.2)^2 = 104.04. \] \[ \text{Variance} = 108 - 104.04 = 3.96. \]
The corrected standard deviation is:
\[ \text{S.D.} = \sqrt{3.96}. \]
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Variance and Standard Deviation Questions
- The mean of 5 observations is \(\frac {24}{5}\) and variance is \(\frac {194}{25}\) . If the mean of first four observations is \(\frac 72\) , then the variance of first four observations is
- Let mean and variance of 6 observations a, b, 68, 44, 40, 60 be 55 and 194. If a > b then find a + 3b
- Given data \(60, 60, 44, 58, 68, α, β, 56\) has mean \(58\), variance = \(66.2\), then find \(α^2 + β^2\).
- From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If the variance of $X$ is $\frac{m}{n}$, where $\gcd(m, n) = 1$, then $n - m$ is equal to ________.
- Let the mean and the standard deviation of the observations $ 2, 3, 4, 5, 7, a, b $ be $ 4 $ and $ \sqrt{2} $ respectively. Then the mean deviation about the mode of these observations is:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,