Let $x_1, x_2, \ldots, x_{10}$ be ten observations such that $\sum_{i=1}^{10}(x_i - 2) = 30$, $\sum_{i=1}^{10}(x_i - \beta)^2 = 98$, $\beta > 2$, and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2(x_1 - 1) + 4\beta$, $2(x_2 - 1) + 4\beta, \ldots, 2(x_{10} - 1) + 4\beta$, then $\frac{\beta\mu}{\sigma^2}$ is equal to:
(1) 100
(2) 120
(3) 110
(4) 90

Marks obtains by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12 . If the number of students whose marks are less than 12 is 18 , then the total number of students is
(1) 52
(2) 48
(3) 44
(4) 40

For a statistical data $x_1, x_2, \ldots, x_{10}$ of 10 values, a student obtained the mean as 5.5 and $\sum_{i=1}^{10} x_i^2 = 371$. He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is
(1) 9
(2) 5
(3) 7
(4) 4

The variance of the numbers $8, 21, 34, 47, \ldots, 320$ is

Q68. Let $\alpha , \beta \in \mathbf { R }$. Let the mean and the variance of 6 observations $- 3,4,7 , - 6 , \alpha , \beta$ be 2 and 23 , respectively. The mean deviation about the mean of these 6 observations is :
(1) $\frac { 13 } { 3 }$
(2) $\frac { 16 } { 3 }$
(3) $\frac { 11 } { 3 }$
(4) $\frac { 14 } { 3 }$

Q68. The frequency distribution of the age of students in a class of 40 students is given below.

Age	15	16	17	18	19	20
No of Students	5	8	5	12	$x$	$y$

If the mean deviation about the median is 1.25 , then $4 x + 5 y$ is equal to :
(1) 46
(2) 43
(3) 44
(4) 47

Q69. If the mean of the following probability distribution of a random variable $X$ :

X	0	2	4	6	8
$\mathrm { P } ( \mathrm { X } )$	$a$	$2 a$	$a + b$	$2 b$	$3 b$

is $\frac { 46 } { 9 }$, then the variance of the distribution is
(1) $\frac { 173 } { 27 }$
(2) $\frac { 566 } { 81 }$
(3) $\frac { 151 } { 27 }$
(4) $\frac { 581 } { 81 }$

Q69. The mean and standard deviation of 20 observations are found to be 10 and 2 . respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12 . The correct standard deviation is
(1) 1.8
(2) 1.94
(3) $\sqrt { 3.96 }$
(4) $\sqrt { 3.86 }$

Q70. Let $B = \left[ \begin{array} { l l } 1 & 3 \\ 1 & 5 \end{array} \right]$ and $A$ be a $2 \times 2$ matrix such that $A B ^ { - 1 } = A ^ { - 1 }$. If $B C B ^ { - 1 } = A$ and $C ^ { 4 } + \alpha C ^ { 2 } + \beta I = O$, then $2 \beta - \alpha$ is equal to
(1) 16
(2) 2
(3) 8
(4) 10

Q86. Let $\mathrm { a } , \mathrm { b } , \mathrm { c } \in \mathrm { N }$ and $\mathrm { a } < \mathrm { b } < \mathrm { c }$. Let the mean, the mean deviation about the mean and the variance of the 5 observations $9,25 , \mathrm { a } , \mathrm { b } , \mathrm { c }$ be 18,4 and $\frac { 136 } { 5 }$, respectively. Then $2 \mathrm { a } + \mathrm { b } - \mathrm { c }$ is equal to $\_\_\_\_$

Consider the 10 observations 2, 3, 5, 10, 11, 13, 15, 21, a and b such that mean of observation in 9 and variance is 34.2. Then the mean deviation about median is (A) 3 (B) 6 (C) 5 (D) 7

Consider 10 data such that their mean is 10 and variance is 2. If one of the data $\alpha$ is removed and new data entry $\beta$ is inserted. Now new mean is 10.1 and new variance is 1.99 then $( \alpha + \beta )$ is equal to
(A) $10 \bar { x } _ { \text {new } }$
(B) $20 \quad \sigma _ { \text {new } } ^ { 2 }$
(C) 1
(D) 2

If the mean and variance of observations $x, y, 12, 14, 4, 10, 2$ is 8 and 16 respectively where $\mathrm{x} > \mathrm{y}$. Then, the value of $3\mathrm{x} - \mathrm{y}$ is
(A) 24 (B) 22 (C) 20 (D) 18

5. For ALL APPLICANTS.

Miriam and Adam agree to relieve the boredom of the school holidays by eating sweets, but their mother insists they limit their consumption by obeying the following rules.

• Miriam eats as many sweets on any day as there have been sunny days during the holiday so far, including the day in question.
• Adam eats sweets only on rainy days. If day $k$ of the holiday is rainy, then he eats $k$ sweets on that day.

For example, if the holiday is eight days long, and begins Rainy, Sunny, Sunny, ... then the tally of sweet consumption might look like this:

Day	1	2	3	4	5	6	7	8	Total
Weather	R	S	S	R	R	S	S	R
Miriam	0	1	2	2	2	3	4	4	18
Adam	1	0	0	4	5	0	0	8	18

In this case, Miriam and Adam eat the same number of sweets in total.
(i) If the holiday has 30 days, 15 of which are sunny and 15 rainy, what arrangement of sunny and rainy days would lead Miriam to eat the greatest number of sweets in total, and what arrangement would lead to the least number? Give the number of sweets that Miriam eats in each case.
(ii) Show that, in the two cases mentioned in part (i), Adam eats the same number of sweets as Miriam.
(iii) Suppose, in a sequence of sunny and rainy days, we arrange to swap a rainy day with a sunny day that immediately follows it. How does the total number of sweets eaten by Miriam change when we make the swap? What about the total number of sweets eaten by Adam?
(iv) If the holiday has 15 sunny days and 15 rainy days, must Miriam and Adam eat the same number of sweets in total? Explain your answer.
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When processing two-dimensional data, one method is to project the data vertically onto a certain line and use that line as a number line to understand the variance of the one-dimensional data formed by the projection points. For the set of two-dimensional data shown in the figure, which line in the following options would result in the smallest variance of the one-dimensional projected data?
(1) $y = 2 x$
(2) $y = - 2 x$
(3) $y = - x$
(4) $y = \frac { x } { 2 }$
(5) $y = - \frac { x } { 2 }$

Classes A and B each have 40 students taking a mathematics exam (total score 100 points). After the exam, classes A and B adjust their scores using $y_1 = 0.8x_1 + 20$ and $y_2 = 0.75x_2 + 25$ respectively, where $x_1, x_2$ represent the original exam scores of classes A and B, and $y_1, y_2$ represent the adjusted scores of classes A and B. The average adjusted scores for both classes are 60 points, with adjusted standard deviations of 16 and 15 points respectively. Select the correct options.
(1) Every student in class A has an adjusted score not lower than their original score
(2) The average original score of class A is higher than that of class B
(3) The standard deviation of original scores in class A is higher than that in class B
(4) If student A from class A has a higher adjusted score than student B from class B, then A's original score is higher than B's original score
(5) If the number of students in class A with adjusted scores below 60 (failing) is greater than the number in class B, then the number of students in class A with original scores below 60 must be greater than in class B

A school's midterm examination has 29 test takers, and all scores are different. After statistics, the scores at the 25th, 50th, 75th, and 95th percentiles are 41, 60, 74, and 92 points respectively. Later, it was discovered that the scores needed adjustment. The scores of the top 15 students with higher scores should each be increased by 5 points, while the remaining students' scores remain unchanged. Assuming the adjusted scores at the 25th, 50th, 75th, and 95th percentiles are $a$, $b$, $c$, and $d$ points respectively, which of the following options is the tuple $(a, b, c, d)$?
(1) $(41,60,74,92)$ (2) $(41,60,74,97)$ (3) $(41,65,79,97)$ (4) $(46,65,79,92)$ (5) $(46,65,79,97)$

A game has a total of 210 players, each holding gems. Among them, 1 player holds 1 gem, 2 players hold 2 gems, and so on, with 20 players holding 20 gems. What is the 90th percentile of the number of gems held by each player?
(1) 16
(2) 17
(3) 18
(4) 19
(5) 20

Divide the 50 positive integers from 1 to 50 equally into groups A and B, with 25 numbers in each group, such that the median of group A is 1 less than the median of group B. How many ways are there to divide them?
(1) $C_{25}^{50}$
(2) $C_{24}^{48}$
(3) $C_{12}^{24}$
(4) $\left(C_{12}^{24}\right)^{2}$
(5) $C_{24}^{48} \cdot C_{12}^{24}$

A city is divided into east and west regions. Each region has a temperature monitoring station. The city's daily maximum temperature (in degrees Celsius) is recorded as the maximum of the daily temperatures from the two regions. The table below shows the distribution of daily maximum temperatures for the east and west regions over a certain month (30 days).

Temperature $t$	$18 \leq t < 24$	$24 \leq t < 30$	$30 \leq t < 36$	$36 \leq t$
East Region (days)	0	11	14	5
West Region (days)	3	12	15	0

Based on the above table, the distribution of the city's daily maximum temperature for the month is shown in the following table.

Temperature $t$	$18 \leq t < 24$	$24 \leq t < 30$	$30 \leq t < 36$	$36 \leq t$
Days	$A$	$B$	$C$	$D$

Select the option that could be the tuple $(A , B , C , D)$.
(1) $( 0,15,15,0 )$
(2) $( 3,12,15,5 )$
(3) $( 0,9,16,5 )$
(4) $( 3,7,15,5 )$
(5) $( 0,12,13,5 )$

A badminton player competes against four opponents: A, B, C, and D, one match each. After the competition, data from these four matches were collected, recording the total number of smashes by each opponent and the average and standard deviation of the time used per smash. The results are shown in the table below. For example, opponent A made 25 smashes in that match, with an average time of 1.2 seconds per smash and a standard deviation of 0.5 seconds.

Opponent	Number of smashes in that match	Average time per smash (seconds)	Standard deviation of time per smash (seconds)
A	25	1.2	0.5
B	14	1.5	0.3
C	20	1.7	0.2
D	30	1.2	0.4

Based on the above, regarding the performance of opponents A, B, C, and D, select the correct options.
(1) C had the highest average time per smash among the four in that match
(2) D spent the most total time on smashing among the four in that match
(3) A's time per smash in that match was the same as D's for every smash
(4) The range of A's smash times in that match is greater than the range of D's smash times in that match
(5) It is impossible for all of B's smash times in that match to be between 1.4 and 1.6 seconds

6. The area of a rectangle is measured to be $5600 \mathrm {~cm} ^ { 2 }$ correct to 2 significant figures.
The width of the rectangle is measured to be 80 cm correct to the nearest centimetre. Which one of the following expressions gives the greatest possible height of the rectangle?
A $\quad 70.5 \mathrm {~cm}$
B $\quad 75 \mathrm {~cm}$
C $\quad \frac { 5650 } { 85 } \mathrm {~cm}$
D $\quad \frac { 5650 } { 80.5 } \mathrm {~cm}$
E $\frac { 5650 } { 75 } \mathrm {~cm}$
F $\quad \frac { 5650 } { 79.5 } \mathrm {~cm}$

18. A group of five numbers are such that:

• their mean is 0
• their range is 20

What is the largest possible median of the five numbers?
A 0
B 4
C $\quad 4 \frac { 1 } { 2 }$
D $\quad 6 \frac { 1 } { 2 }$
E 8 F 20

There are two sets of data: the mean of the first set is 15 , and the mean of the second set is 20 .
One of the pieces of data from the first set is exchanged with one of the pieces of data from the second set.
As a result, the mean of the first set of data increases from 15 to 16, and the mean of the second set of data decreases from 20 to 17.
What is the mean of the set made by combining all the data?
A $16 \frac { 1 } { 4 }$
B $16 \frac { 1 } { 3 }$
C $16 \frac { 1 } { 2 }$
D $\quad 16 \frac { 2 } { 3 }$
E $16 \frac { 3 } { 4 }$

A set of six distinct integers is split into two sets of three. The first set of three integers has a mean of 10 and a median of 8 . The second set of three integers has a mean of 12 and a median of 9 . What is the smallest possible range of the set of all six integers?
A 8
B 10
C 11
D 12
E 14 F 15