The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is _______.
Correct Answer: 96
Solution and Explanation
Given that,
np + npq = 82.5 … (1)
np (npq) = 1350 … (2)
Since, Mean and Vairance be the roots of x2 – 82.5x + 1350 = 0
⇒ x2 – 22.5 x – 60x + 1350 = 0
⇒ x – (x – 22.5) – 60 (x – 22.5) = 0
Mean = 60 and Variance = 22.5
np = 60, npq = 22.5
\(⇒q=\frac{9}{24}=\frac{3}{8},\)
\(⇒p=\frac{5}{8}\)
\(∴n\frac{5}{8}=60\)
\(⇒n=96\)
So, the answer is 96.
Top Questions on binomial distribution
- The mean and variance of a binomial variate \(X\) are \(\frac{16}{5}\) and \(\frac{48}{25}\) respectively. Given that \(P(X \geq 1) = 1 - K \left(\frac{3}{5}\right)^7\), find \(5K\):
- TS EAMCET - 2024
- Mathematics
- binomial distribution
- The mean and variance of a random variable \( X \) having a Binomial distribution are 4 and 2 respectively, then \( P(X=1) \) is
- TS PGECET - 2024
- Engineering Mathematics
- binomial distribution
- In a binomial distribution, the mean is 4 and variance is 3. Then, its mode is:
- BITSAT - 2024
- Mathematics
- binomial distribution
- If \( X \sim B(5, p) \) is a binomial variate such that \( p(X = 3) = p(X = 4) \), then \( P(|X - 3| < 2) = \dots \)
- AP EAPCET - 2024
- Mathematics
- binomial distribution
- In a Binomial distribution, the difference between the mean and standard deviation is 3, and the difference between their squares is 21. Then, the ratio \( P(x = 1) : P(x = 2) \) is:
- AP EAMCET - 2024
- Mathematics
- binomial distribution
Questions Asked in JEE Main exam
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
- JEE Main - 2025
- Solutions
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.- JEE Main - 2025
- Electrical Circuits
- Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k} \). Let \( L_1 : \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \vec{a}, \lambda \in {R} \) and \( L_2 : \vec{r} = (\hat{j} + \hat{k}) + \mu \vec{b}, \mu \in \mathbb{R} \) be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \vec{a} + \vec{b} \), then \( L_3 \) passes through the point:
- JEE Main - 2025
- Vectors
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
Concepts Used:
Mean Value Theorem
The theorem states that for a curve f(x) passing through two given points (a, f(a)), (b, f(b)), there is at least one point (c, f(c)) on the curve where the tangent is parallel to the secant passing via the two given points is the mean value theorem.
The mean value theorem is derived herein calculus for a function f(x): [a, b] → R, such that the function is continuous and differentiable across an interval.
- The function f(x) = continuous across the interval [a, b].
- The function f(x) = differentiable across the interval (a, b).
- A point c exists in (a, b) such that f'(c) = [ f(b) - f(a) ] / (b - a)