x | 5 | 10 | 15 | 20 | 25 |
f | 6 | 8 | 6 | y | 5 |
Given: The mean of the frequency distribution is 15.
Step 1: Understanding the Mean Formula
The mean is given by:
\[ \text{Mean} = \frac{\sum fx}{\sum f} \]
Where:
Step 2: Calculating \( \sum fx \) and \( \sum f \)
Calculating \( fx \) for given values:
\[ (5 \times 6) + (10 \times 8) + (15 \times 6) + (20 \times y) + (25 \times 5) \]
\[ = 30 + 80 + 90 + 20y + 125 \]
\[ \sum fx = 325 + 20y \]
Calculating \( \sum f \):
\[ \sum f = 6 + 8 + 6 + y + 5 = 25 + y \]
Step 3: Substituting into Mean Formula
\[ 15 = \frac{325 + 20y}{25 + y} \]
Multiplying both sides by \( (25 + y) \):
\[ 15(25 + y) = 325 + 20y \]
\[ 375 + 15y = 325 + 20y \]
Rearranging:
\[ 375 - 325 = 20y - 15y \]
\[ 50 = 5y \]
\[ y = 10 \]
Final Answer: \( \mathbf{10} \)
To find the value of y in the frequency distribution table, given that the mean is 15, we start by applying the formula for the mean of a grouped frequency distribution, which is:
Mean = Σ(fi×xi)/Σfi
Given:
x | 5 | 10 | 15 | 20 | 25 |
f | 6 | 8 | 6 | y | 5 |
Calculate Σ(fi×xi):
So, Σ(fi×xi) = 30 + 80 + 90 + 20y + 125 = 325 + 20y
Next, calculate Σfi:
6 + 8 + 6 + y + 5 = 25 + y
Using the mean formula, substitute the values:
15 = (325 + 20y)/(25 + y)
\(\Rightarrow\) Cross multiply to solve for y:
\(\Rightarrow\) 15(25 + y) = 325 + 20y
\(\Rightarrow\) 375 + 15y = 325 + 20y
Rearrange to solve for y:
\(\Rightarrow\) 375 - 325 = 20y - 15y
\(\Rightarrow\) 50 = 5y
Divide both sides by 5:
\(\Rightarrow\) y = 10
The value of y is 10.
Number of students per Teacher | Number of Schools |
20 - 25 | 5 |
25 - 30 | 15 |
30 - 35 | 25 |
35 - 40 | 30 |
40 - 45 | 15 |
45 - 50 | 10 |
Class Interval | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
---|---|---|---|---|---|
Number of Students | 15 | 18 | 21 | 29 | 17 |
Number of students per Teacher | Number of Schools |
20 - 25 | 5 |
25 - 30 | 15 |
30 - 35 | 25 |
35 - 40 | 30 |
40 - 45 | 15 |
45 - 50 | 10 |
x | Less than 10 | Less than 20 | Less than 30 | Less than 40 | Less than 50 | Less than 60 |
f | 3 | 12 | 27 | 57 | 75 | 80 |
If the variance of the frequency distribution
xi | Frequency ft |
2 | 3 |
3 | 6 |
4 | 16 |
5 | \(\alpha\) |
6 | 9 |
7 | 5 |
8 | 6 |
is 3 , then $\alpha$ is equal to