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
\(\frac 32\)
\(\frac 52\)
\(\frac 54\)
\(\frac 23\)
The Correct Option is C
Solution and Explanation
The correct option is (C): \(\frac 54\).
Top Questions on Variance and Standard Deviation
- 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 the correct standard deviation are $ \mu $ and $ \sigma $ respectively, then $ 10(\mu + \sigma) $ is equal to:
- JEE Main - 2025
- Mathematics
- Variance and Standard Deviation
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then \(a + b + ab\) is equal to:
- JEE Main - 2025
- Mathematics
- Variance and Standard Deviation
- 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:
- JEE Main - 2025
- Mathematics
- Variance and Standard Deviation
- Variance of 240, 260, 270, 280
- KEAM - 2025
- Mathematics
- Variance and Standard Deviation
- Find the standard deviation of the numbers: -3, 0, 3, 8.
- KEAM - 2025
- Mathematics
- Variance and Standard Deviation
Questions Asked in JEE Main exam
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____.
- JEE Main - 2025
- Magnetic Field
- Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.
- JEE Main - 2025
- Solutions
- A physical quantity \( Q \) is related to four observables \( a \), \( b \), \( c \), and \( d \) as follows: \[ Q = \frac{a b^4}{c d^2} \] Where:
- \( a = (60 \pm 3) \, \text{Pa} \),
- \( b = (20 \pm 0.1) \, \text{m} \),
- \( c = (40 \pm 0.2) \, \text{N·s/m}^2 \),
- \( d = (50 \pm 0.1) \, \text{m} \).
Then the percentage error in \( Q \) is:- JEE Main - 2025
- Error analysis
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: - Assume a living cell with 0.9%(\(w/w\)) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data up to first decimal place only) The cell will:
- JEE Main - 2025
- osmosis
Concepts Used:
Variance and Standard Deviation
Variance:
According to layman’s words, the variance is a measure of how far a set of data are dispersed out from their mean or average value. It is denoted as ‘σ2’.
Variance Formula:
Read More: Difference Between Variance and Standard Deviation
Standard Deviation:
The spread of statistical data is measured by the standard deviation. Distribution measures the deviation of data from its mean or average position. The degree of dispersion is computed by the method of estimating the deviation of data points. It is denoted by the symbol, ‘σ’.
Types of Standard Deviation:
- Standard Deviation for Discrete Frequency distribution
- Standard Deviation for Continuous Frequency distribution
Standard Deviation Formulas:
1. Population Standard Deviation
2. Sample Standard Deviation
