Statistics Set-2

Q1. A class of 40 students has mean marks $72$ and another class of 35 students has mean marks $68$. The mean marks of the combined group of $75$ students is

Q2. The following frequency distribution gives heights (in cm) of 30 students: classes $0$–$10$: $4$, $10$–$20$: $10$, $20$–$30$: $12$, $30$–$40$: $4$. Using the formula for median of grouped data, the median height is

Q3. The mean marks of 50 students in a test is $72$. After re-evaluation, each of the top 10 students is awarded $5$ extra marks and each of the remaining 40 students loses $2$ marks. The new mean is

Q4. Assertion (A): For any dataset, the median is the value that minimizes the sum of absolute deviations from it.
Reason (R): The arithmetic mean is the value that minimizes the sum of squared deviations from it.

Q5. A frequency distribution is given by classes and frequencies: $0$–$10$: $8$, $10$–$20$: $12$, $20$–$30$: $x$, $30$–$40$: $10$, $40$–$50$: $5$. If the median of the distribution is $24$, the missing frequency $x$ equals

Q6. Two sections of a school have averages of marks as follows: Section I with 30 students has mean $65$ and Section II with 20 students has mean $70$. What is the combined mean of both sections?

Q7. A discrete data set has values $10,20,30,40$ with corresponding frequencies $3,4,x,2$ respectively. If the mean of the distribution is $26$, the missing frequency $x$ equals

Q8. For the following grouped frequency distribution, find the median (to two decimal places).
Classes: $0$–$10$ : $5$, $10$–$20$ : $9$, $20$–$30$ : $15$, $30$–$40$ : $6$.

Q9. Consider the grouped frequency distribution with equal class width $10$:
Classes $0$–$10:8$, $10$–$20:12$, $20$–$30:12$, $30$–$40:5$. If a tie for highest frequency is resolved by taking the first of the tied classes ($10$–$20$) as the modal class, the mode (by using the formula

) is

Q10. Assertion (A): If a dataset of $N$ observations has mean $m$, then adding one new observation equal to $m$ does not change the mean.
Reason (R): Let the original sum be $S=Nm$. Adding $m$ gives new sum $S+m=Nm+m=m(N+1)$, and dividing by $N+1$ yields the same mean $m$.

Q11. The following frequency distribution gives the marks obtained by $20$ students in a test: $10$–$14$: $2$, $15$–$19$: $5$, $20$–$24$: $7$, $25$–$29$: $6$. Find the mean mark $\bar{x}$ of the distribution.

Q12. A data set of $10$ observations is: $6,8,5,10,9,7,8,6,11,5$. Compute the population variance $\sigma^2=\dfrac{1}{n}\sum (x_i-\bar{x})^2$.

Q13. For the following grouped frequency distribution, use the formula for mode $\displaystyle \text{Mode}=l+\frac{(f_1-f_0)}{(2f_1-f_0-f_2)}\times h$ to find the modal value. Classes: $10$–$20$: $5$, $20$–$30$: $12$, $30$–$40$: $20$, $40$–$50$: $8$. (Here $l$ is lower limit of modal class, $h$ is class width.)

Q14. Assertion (A): For a perfectly symmetric distribution of data, the mean equals the median.
Reason (R): The arithmetic mean is obtained by summing all observations and dividing by the number of observations.
Choose the correct option:

Q15. A grouped frequency distribution is given as: $0$–$10$: $8$, $10$–$20$: $12$, $20$–$30$: $8$, $30$–$40$: $12$. Find the median of the data using the grouped median formula $\displaystyle \text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h$, where $cf$ is cumulative frequency before the median class.