# RRB NTPC Elementary Statistics Questions PDF

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## RRB NTPC Elementary Statistics Questions PDF

Download RRB NTPC Elementary Statistics Questions and Answers PDF. Top 20 RRB NTPC Maths questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: What is the relation among mean, median & mode ?

a) mode – 2(mean) = 3(median)

b) mode + 3(mean) = 2(median)

c) mode + 2(mean) = 3(median)

d) mode – 3(mean) = 2(median)

Question 2: Find the mode if mean and median are 4 & 5 respectively ?

a) 5

b) 7

c) 9

d) 11

Question 3: In a moderately skewed distribution, it is known that the median is 5 and the mode is 7. What is the mean?

a) 10

b) 8

c) 6

d) 4

Question 4: Find the mode of 2,3,4, 2,2,3,2,3,4,5,3,5,3,2,5,2 ?

a) 2

b) 3

c) 4

d) 5

Question 5: Find the mode of 2, 3, 2, 2, 3, 4, 5, 4, 3, 2, 3, 5, 3 ?

a) 2

b) 3

c) 3.5

d) 2.5

Question 6: In a moderately symmetric distribution, it is known that the median and mean are 6 and 4 respectively. Find the mode of the distribution.

a) 8

b) 10

c) 6

d) 12

Question 7: The arithmetic mean of 9 distinct integers is 87. If none of the numbers is more than 100 and the average of the smallest five numbers is 78, find the minimum value of the sixth number.

Question 8: What is the minimum value of the function $f(x)= x^4-8x^3+22x^2-24x+4$?

a) $- \infty$

b) $-5$

c) $-\frac{25}{4}$

d) None of these

Question 9: If a, b, and c are real numbers such that a+b+c=30 and ab+bc+ca=192. Find the maximum value of a.

a) 30

b) 22

c) 29

d) None of these

Question 10: Find the difference of mean and median of the following data set {2,3,4,5,5,8,8}

a) 1

b) 2

c) 3

d) 0

Question 11: A teacher observes the marks of the five students of the class. She observes the median and mean each to be 5 but the only mode is 8. What is the lowest marks obtained?

a) 1

b) 2

c) 3

d) Can’t be determined

Question 12: The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the largest of the six integers is

a) 26

b) 28

c) 30

d) 32

e) 34

Question 13: What is the difference between the mean and median of set S = {2, 4, 6, 7, 7, 13, 18, 92}?

a) 10.125

b) 11.125

c) 11.625

d) 14

e) 10.875

Question 14: A teacher noticed a strange distribution of marks in the exam. There were only three distinct scores: 6, 8 and 20. The mode of the distribution was 8. The sum of the scores of all the students was 504. The number of students in the in most populated category was equal to the sum of the number of students with lowest score and twice the number of students with the highest score. The total number of students in the class was:

a) 50

b) 51

c) 57

d) 56

Question 15: Consider the set of numbers 11, 3, 6, 3, 5, 3 and x. If the mean, median and mode of this set of numbers are in an non-constant arithmetic progression. What are the number of possible values for x?

a) 0

b) 1

c) 2

d) None of the above

Question 16: What is a if the mean of the set S={4, 10, a, 12, 13, 21} is same as the mode?

a) 10

b) 10.75

c) 12

d) Cannot be determined

Question 17: What is the difference between the median and mode of S={1, 2, 4, 4, 8, 14, 32, 64}?

a) 0

b) 2

c) 4

d) Either 0 or 4

Question 18: What is the difference in the median and mode of the list { 6,2,3,4,3,5,2,7,4,2,6}

a) 1

b) 2

c) 3

d) 2.5

Question 19: What is the median of the set of all three digit natural numbers such that the units digit is greater than the tens digit which is greater than the hundred’s digit?

a) 263

b) 258.5

c) 345

d) 278.5

Question 20: What is the difference between the mean and median of set S={2, 4, 6, 7, 7, 13, 18, 92}?

a) 10.125

b) 11.125

c) 11.625

d) 14

Mode = 3(Median)-2(Mean)

Mode+2(Mean) = 3(Median)

So the answer is option C.

Mode = 3(Median)-2(Mean)

Mode = 3(5)-2(4) = 15 – 8 = 7

So the answer is option B.

Mode = 3(Median)-2(Mean)

7 = 3(5)-2(mode)

7-15 = -2(mode)

-8 = -2(mode)

mode = 4

So the answer is option D.

Mode is the number which occurred frequently = 2

So the answer is option A.

Mode is the most frequently occurred number = 3

So the answer is option B.

$3(median)-2(mean) = mode$

$3(6)-2(4) = mode$

$18-8 = mode$

$10 = mode$

So the answer is option B.

If the average of 9 numbers is 87, then the sum of these 9 distinct numbers will be 9 X 87 = 783
Let the numbers be a1, a2, a3, …..a9 where a9> a8> …a1.
So, a1+a2+a3+a4+a5 = 78 X 5 = 390
Smallest value of a5 can be 80 when a1, a2, a3, a4, and a5 are 76, 77, 78, 79, 80.
This means that a6 > 80.
Now, the sum of the rest of the four numbers is 738-390 = 393
For a6 to be min, a7, a8, a9 must be max.
=> a6 + 98 + 99 + 100 = 393 or a6 = 96.
Thus, 96 is the correct answer.

$f(x)= x^4-8x^3+22x^2-24x+4$
$f'(x)=4x^3-24x^2+44x-24$
and $f”(x)=12x^2-48x+44$
f’(x)=0
=> $4x^3-24x^2+44x-24=0$ or $(x-1)(x-2)(x-3)=0$
The critical points are x=1,2,3
Putting these points in f’’(x) we get f(1)=+ve, f(2)=-ve, f(3)=+ve
So, at x=1 and 3 there is local minima and x=2 there is local maxima. This means, the graph would be increasing beyond x=1 and x=3.

Thus, the lower value between f(1) and f(3) will be the minimum value of f(x)
Both f(1)=f(3)=-5
Thus, minimum value is -5.

a+b+c=30 and ab+bc+ca=192
=> b(a+c)+ca=192
b=30-(a+c)
=> [30-(a+c)](a+c)+ac=192
=> $30(a+c)-(a+c)^2+ca=192$
=> $30a+30c-a^2-c^2-ca=192$
=> $c^2+(a-30)c+(a^2-30a+192)=0$
Now, c is real. This means that discriminant of the above equation in c should be $\geq0$.
=> $(a-30)^2-4(a^2-30a+192)\geq0$
=>$a^2+900-60a-4a^2+120a-4X192\geq0$
=> $a^2+900-60a-4a^2+120a-4X192\geq0$
=> $3a^2-60a-132\leq0$
=> $a^2-20a-44\leq0$
=> $(a-22)(a+2)\leq0$
=> $a-22\geq 0=a\geq22$ and $a+2\leq 0=a\leq-2$, which is not possible
Or $a-22\leq 0=a\leq22$ and $a+2\geq 0=a\geq-2$
Thus, max possible value is 22.

Median is the middle term which is 5.
Mean=(2+3+4+5+5+8+8)/7=5
Hence difference=0

Let the marks of the students in ascending order be a,b,c,d,e
a+b+c+d+e=5*5=25
Median=5
c=5
Mode=8
d and e each is equal to 8 because if there are more than 2 8’s the sum becomes more than 25
a+b+5+8+8=25 a+b=4
The answers can be a=1 and b=3 or a=0 and b=4
Hence, the answer is can’t be determined. a=b=2 cannot be possible because there is only one mode

Since the mean is 18, the total is 90. Let the numbers be a, b, c, d, e and f in ascending order. a+b+c+d+e+f=90

The median is 18, so there are three numbers less than 18 and three numbers more than 18. Therefore, a,b,c <18 and d,e,f >18. The median is the average of c and d. Therefore, (c+d)/2=18 => c+d=36

Here we need to maximize f, so we need to minimize a, b, c, d and e.

The minimum value a and b can take is 1, the least positive integer.

To minimize e, e should be equal to d+1. It cannot be equal to d as the mode is less than 18.

Hence, the minimum value that d can take such that c+d=36 and d>18 is d=19. Therefore, c=17 and e=20

Hence, a = 1, b = 1, c = 17, d = 19, e = 20. Therefore, f = 90 -(1+1+17+19+20) = 32.

Sum of the numbers = 2+4+6+7+7+13+18+92 = 149

So, mean = 149/8 = 18.625

Median = 7 (frequency of occurrence = 2)

Hence, difference = 18.625 – 7 = 11.625

6a+8b+20c = 504 and b = a+2c

The only integral solution to the above equation is when a = 18 b = 32 and c = 7.

Hence, the total number of students in the class were 57

It is given mean,median and mode are in A.P. Hence we have mean + mode = 2*median
The mode of the set of numbers is 3. And the mean of the numbers is (x+31)/7.
If x<3, the median is 3. But, as the arithmetic progression is non constant, no such values are possible.
If 3 < x < 5, the median is x. In this case, (x+31)/7+3 = 2x or x = 4. This lies in between 3 and 5 and is a valid solution.
If x > 5, the median is 5. So, (x+31)/7 + 3 = 10 or x = 18. As this is greater than 5, this is also a valid solution.
So, total number of valid solutions is 2

Let m be the mode and M the mean. Hence, M=(60+a)/6=10+a/6.
For the set to have a mode, a must be one of the five elements already in S i.e. 4, 10,12, 13 and 21.
As M is 10+a/6 it cannot be 4 or 10. If a=12,13,21 => M = 12, 12.167, 13.5.
Hence only for a=12 is the mean and mode same. Hence a=12.

Mode=4, Median = (4+8)/2=6. Hence difference=2

The list when ordered in ascending order is {2,2,2,3,3,4,4,5,6,6,7}.
So, the median is 4 and the mode is 2. Hence, difference is 2

Let the three digit number be a b c. Hence, a<b<c. If a=1, b=2 c=3-9 = 7 numbers.
Similarly, a=1, b=2, c=4-9 (6 numbers). Hence # of numbers starting with 1 = (7+6+5+4+3+2+1) =28.
Starting with 2 = (6+5+..+1) = 21. Hence, the number of numbers are = 28+21+15+10+6+3+1=84.
Hence median is average of 42nd and 43rd term. 28+21= 49 terms are less than 300.
Hence 289-49, 279-48, 278-47, 269-46, 268 – 45, 267-44, 259-43 and 258-42.
Hence 42nd term is 258 and 43rd is 259. Hence median is 258.5.

The mean of the set is the average of all the numbers in the set = (2 + 4 + 6 + 7 + 7 + 13 + 18 + 92)/8 = 18.625
The median of the set is the middle number of the set when it is placed in ascending order. In this case, it is 7.
Hence, the difference is 11.625

We hope this Elementary Statistics questions for RRB NTPC Exam will be highly useful for your preparation.