The frequency distribution of the age of students in a class of 40 students is given below:\(Age\) 15 16 17 18 19 20 No. of Students 5 8 5 12 x y
If the mean deviation about the median is 1.25, then \(4x + 5y\) is equal to:
| \(Age\) | 15 | 16 | 17 | 18 | 19 | 20 |
|---|---|---|---|---|---|---|
| No. of Students | 5 | 8 | 5 | 12 | x | y |
If the mean deviation about the median is 1.25, then \(4x + 5y\) is equal to:
- 43
- 44
- 47
- 46
The Correct Option is B
Approach Solution - 1
The problem is to find the value of \(4x + 5y\) using the given frequency distribution and information about the mean deviation around the median.
The age distribution given in the question is:
| \(Age\) | 15 | 16 | 17 | 18 | 19 | 20 |
|---|---|---|---|---|---|---|
| No. of Students | 5 | 8 | 5 | 12 | x | y |
The total number of students is:
\(5 + 8 + 5 + 12 + x + y = 40\)
So, we have:
\(x + y = 10\)
Next, we need to find the median age. The median is the middle value in a frequency distribution when the students are arranged in ascending order. To find the median, we need the cumulative frequency distribution:
| \(Age\) | 15 | 16 | 17 | 18 | 19 | 20 |
|---|---|---|---|---|---|---|
| Cumulative Frequency | 5 | 13 | 18 | 30 | 30+x | 40 |
The median corresponds to the 20th value since the total number of students is 40. From the cumulative distribution, we can see that the median age is 18.
Now, the mean deviation about the median is given by:
\(\frac{1}{N}\sum |x_i - Median|\)
With a mean deviation of 1.25, we have:
\(\frac{1}{40}[5|15-18| + 8|16-18| + 5|17-18| + 12|18-18| + x|19-18| + y|20-18|] = 1.25\)
Simplifying within the modulus:
\(\frac{1}{40}(15 + 16 + 5 + 0 + x + 2y) = 1.25\)
Simplifying further:
\(36 + x + 2y = 50\)
Which simplifies to:
\(x + 2y = 14\)
We now have two equations:
- \(x + y = 10\)
- \(x + 2y = 14\)
Subtract the first equation from the second:
\((x + 2y) - (x + y) = 14 - 10\)
Solving gives:
\(y = 4\)
Substitute \(y = 4\) in \(x + y = 10\):
\(x + 4 = 10\)
\(x = 6\)
Therefore, \(4x + 5y\) becomes:
\(4(6) + 5(4) = 24 + 20 = 44\)
Thus, the value of \(4x + 5y\) is 44.
Approach Solution -2
We are given:
x + y = 10 $\quad \cdots \text{(1)}$
The median is:
M = 18
The formula for the Mean Deviation (M.D.) is:
$\text{M.D.} = \frac{\sum f_i |x_i - M|}{\sum f_i}$
Substituting the given values:
1.25 = $\frac{36 + x + 2y}{40}$
Simplifying:
x + 2y = 14 $\quad \cdots \text{(2)}$
From equations (1) and (2), solving simultaneously:
x + y = 10
x + 2y = 14
Subtracting (1) from (2):
y = 4
Substituting y = 4 into (1):
x = 6
Now, substituting x = 6 and y = 4 into 4x + 5y:
4x + 5y = 4(6) + 5(4) = 24 + 20 = 44
Final Answer: 44
The table values are as follows:
| Age \(x_i\) | f | \(|x_i - M|\) | \(f_i|x_i - M|\) |
|---|---|---|---|
| 15 | 3 | 3 | 15 |
| 16 | 8 | 2 | 16 |
| 17 | 5 | 1 | 5 |
| 18 | 12 | 0 | 0 |
| 19 | x | 1 | x |
| 20 | y | 2 | 2y |
Top Questions on Statistics
Let the Mean and Variance of five observations $ x_i $, $ i = 1, 2, 3, 4, 5 $ be 5 and 10 respectively. If three observations are $ x_1 = 1, x_2 = 3, x_3 = a $ and $ x_4 = 7, x_5 = b $ with $ a>b $, then the Variance of the observations $ n + x_n $ for $ n = 1, 2, 3, 4, 5 $ is
- JEE Main - 2025
- Mathematics
- Statistics
- Marks obtained by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is:
- JEE Main - 2025
- Mathematics
- Statistics
Find the variance of the following frequency distribution:
Class Interval 0--4 4--8 8--12 12--16 Frequency 1 2 2 1 - AP EAPCET - 2025
- Mathematics
- Statistics
- The variance of the discrete data 3, 4, 5, 6, 7, 8, 10, 13 is
- TS EAMCET - 2025
- Mathematics
- Statistics
- Suppose a, b and c are three real numbers such that Max(a, b, c) + Min(a, b, c) = 15, and Median(a, b, c) - Mean(a, b, c) = 2. Then the median of a, b and c is
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