Class 5-10 10-15 15-20 20-25 25-30 30-35 Frequency 2 k 28 54 k+1 5
is 21, then k is one of the roots of the equation :
Calculate Mean Components
To find the mean of the frequency distribution, we first need the midpoint ($x_i$) of each class interval. We list these values and the corresponding frequencies ($f_i$) in a table.
| Class | Frequency ($f_i$) | Midpoint ($x_i$) | Product ($f_i \cdot x_i$) |
| 5-10 | 2 | $ \frac{5+10}{2} = 7.5 $ | $ 2 \times 7.5 = 15 $ |
| 10-15 | $k$ | $ \frac{10+15}{2} = 12.5 $ | $ k \times 12.5 = 12.5k $ |
| 15-20 | 28 | $ \frac{15+20}{2} = 17.5 $ | $ 28 \times 17.5 = 490 $ |
| 20-25 | 5 | $ \frac{20+25}{2} = 22.5 $ | $ 5 \times 22.5 = 112.5 $ |
| 25-30 | $4k+1$ | $ \frac{25+30}{2} = 27.5 $ | $ (4k+1) \times 27.5 = 110k + 27.5 $ |
| 30-35 | 5 | $ \frac{30+35}{2} = 32.5 $ | $ 5 \times 32.5 = 162.5 $ |
Formulate Mean Equation
Next, we calculate the sum of all frequencies ($\sum f_i$) and the sum of the products of frequencies and midpoints ($\sum f_i x_i$).
- Sum of Frequencies: $ \sum f_i = 2 + k + 28 + 5 + (4k+1) + 5 = 41 + 5k $
- Sum of Products: $ \sum f_i x_i = 15 + 12.5k + 490 + 112.5 + (110k + 27.5) + 162.5 = 807.5 + 122.5k $
The mean ($\bar{x}$) for grouped data is calculated using the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$. We are given that the mean is 21.
$ 21 = \frac{807.5 + 122.5k}{41 + 5k} $Determine the Root 'k'
The question states that $k$ is a root of the correct quadratic equation. Let's find the roots of the equation given in Option C: $2x^2 - 19x - 10 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$ x = \frac{-(-19) \pm \sqrt{(-19)^2 - 4(2)(-10)}}{2(2)} $ $ x = \frac{19 \pm \sqrt{361 + 80}}{4} $ $ x = \frac{19 \pm \sqrt{441}}{4} $ $ x = \frac{19 \pm 21}{4} $This gives two possible roots:
- $ x_1 = \frac{19 + 21}{4} = \frac{40}{4} = 10 $
- $ x_2 = \frac{19 - 21}{4} = \frac{-2}{4} = -\frac{1}{2} $
In the context of a frequency distribution, the frequency ($k$) cannot be negative. Therefore, the relevant value for $k$ is $10$. Since $k=10$ is a root of the equation $2x^2 - 19x - 10 = 0$, Option C is the correct equation.
Similar Questions
A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn
at random is tossed and head turns up. If the probability that the drawn coin was unbiased,
is $\frac{m}{n}$, gcd (m, n) = 1, then $n^2 - m^2$ is equal to :
Given three indentical bags each containing 10 balls, whose colours are as follows :
Red Blue Green Bag I 3 2 5 Bag II 4 3 3 Bag III 5 1 4 A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is $p$ and if the ball is Green, the probability that it is from bag III is $q$, then the value of $\left( \frac{1}{p} + \frac{1}{q} \right)$ is:
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- From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is
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- A variable X takes values $0, 0, 2, 6, 12, 20, \dots, n(n-1)$ with frequencies ${}^{n}C_{0}, {}^{n}C_{1}, {}^{n}C_{2}, {}^{n}C_{3}, {}^{n}C_{4}, {}^{n}C_{5}, \dots, {}^{n}C_{n}$, respectively. If the mean of this data is 60, then its median is :
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Important Questions from Statistics and Probability
A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn
at random is tossed and head turns up. If the probability that the drawn coin was unbiased,
is $\frac{m}{n}$, gcd (m, n) = 1, then $n^2 - m^2$ is equal to :
Given three indentical bags each containing 10 balls, whose colours are as follows :
Red Blue Green Bag I 3 2 5 Bag II 4 3 3 Bag III 5 1 4 A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is $p$ and if the ball is Green, the probability that it is from bag III is $q$, then the value of $\left( \frac{1}{p} + \frac{1}{q} \right)$ is:
- From the first 100 natural numbers, two numbers first a and then b are selected randomly without replacement. If the probability that $a - b \ge 10$ is $\frac{m}{n}$, gcd (m, n) = 1, then $m + n$ is equal to ___________.
- From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is
- The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is :
