Class 0−44−88−1212−1616−20Frequency 391086Cumulative frequency 312223036 M=l+(f2N−C)h
Here, N=3+9+10+8+6=36
⇒2N=18
So, median class is (8−12)
⇒l=8,C=22,f=10,h=4
⇒M=8+1018−12×4
⇒M=10.4
⇒20M=208
JEE Main 2024 — Mathematics Probability & Statistics
Let M denote the median of the following frequency distribution.
Class Frequency 0−434−898−121012−16816−206
Then 20M is equal to :
Held on 30 Jan 2024 · Verified 6 Jul 2026.
416
104
52
208
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If a random variable $x$ has the probability distribution $\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(x) & 0 & 2k & k & 3k & 2k^{2} & 2k & k^{2}+k & 7k^{2} \\ \hline \end{array}$ then $P(3 < x \leq 6)$ is equal to
If the mean of the data <table class="pyq-table"><tbody><tr><th>Class</th><th>$5-10$</th><th>$10-15$</th><th>$15-20$</th><th>$20-25$</th><th>$25-30$</th><th>$30-35$</th></tr><tr><td>Frequency</td><td>$2$</td><td>$k$</td><td>$28$</td><td>$54$</td><td>$k+1$</td><td>$5$</td></tr></tbody></table> is $21$, then $k$ is one of the roots of the equation :
A bag contains $(N+1)$ coins $- N$ fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\dfrac{9}{16}$, then $N$ is equal to:
Let the mean and the variance of seven observations $2, 4, \alpha, 8, \beta, 12, 14$, $\alpha < \beta$, be $8$ and $16$ respectively. Then the quadratic equation whose roots are $3\alpha + 2$ and $2\beta + 1$ is :
If the mean and the variance of the data \(\begin{array}{|c|c|c|c|c|} \hline \text{Class} & 4\text{-}8 & 8\text{-}12 & 12\text{-}16 & 16\text{-}20 \\ \hline \text{Frequency} & 3 & \lambda & 4 & 7 \\ \hline \end{array}\) are $\mu$ and 19 respectively, then the value of $\lambda+\mu$ is
Work through every JEE Main Probability & Statistics PYQ, year by year.