{"id":26785,"date":"2019-03-29T18:28:17","date_gmt":"2019-03-29T12:58:17","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=26785"},"modified":"2019-04-01T16:01:48","modified_gmt":"2019-04-01T10:31:48","slug":"rrb-ntpc-elementary-statistics-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/rrb-ntpc-elementary-statistics-questions-pdf\/","title":{"rendered":"RRB NTPC Elementary Statistics Questions PDF"},"content":{"rendered":"

RRB NTPC Elementary Statistics Questions PDF<\/strong><\/span><\/h2>\n

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.<\/p>\n

Download RRB NTPC Elementary Statistics<\/a><\/p>\n\n

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Question 1:\u00a0<\/b>What is the relation among mean, median & mode ?<\/p>\n

a)\u00a0mode – 2(mean) = 3(median)<\/p>\n

b)\u00a0mode + 3(mean) = 2(median)<\/p>\n

c)\u00a0mode + 2(mean) = 3(median)<\/p>\n

d)\u00a0mode – 3(mean) = 2(median)<\/p>\n

Question 2:\u00a0<\/b>Find the mode if mean and median are 4 & 5 respectively ?<\/p>\n

a)\u00a05<\/p>\n

b)\u00a07<\/p>\n

c)\u00a09<\/p>\n

d)\u00a011<\/p>\n

Question 3:\u00a0<\/b>In a moderately skewed distribution, it is known that the median is 5 and the mode is 7. What is the mean?<\/p>\n

a)\u00a010<\/p>\n

b)\u00a08<\/p>\n

c)\u00a06<\/p>\n

d)\u00a04<\/p>\n

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Question 4:\u00a0<\/b>Find the mode of 2,3,4, 2,2,3,2,3,4,5,3,5,3,2,5,2 ?<\/p>\n

a)\u00a02<\/p>\n

b)\u00a03<\/p>\n

c)\u00a04<\/p>\n

d)\u00a05<\/p>\n

Question 5:\u00a0<\/b>Find the mode of 2, 3, 2, 2, 3, 4, 5, 4, 3, 2, 3, 5, 3 ?<\/p>\n

a)\u00a02<\/p>\n

b)\u00a03<\/p>\n

c)\u00a03.5<\/p>\n

d)\u00a02.5<\/p>\n

Question 6:\u00a0<\/b>In a moderately symmetric distribution, it is known that the median and mean are 6 and 4 respectively. Find the mode of the distribution.<\/p>\n

a)\u00a08<\/p>\n

b)\u00a010<\/p>\n

c)\u00a06<\/p>\n

d)\u00a012<\/p>\n

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Question 7:\u00a0<\/b>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.<\/p>\n

Question 8:\u00a0<\/b>What is the minimum value of the function $f(x)= x^4-8x^3+22x^2-24x+4$?<\/p>\n

a)\u00a0$- \\infty$<\/p>\n

b)\u00a0$-5$<\/p>\n

c)\u00a0$-\\frac{25}{4}$<\/p>\n

d)\u00a0None of these<\/p>\n

Question 9:\u00a0<\/b>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.<\/p>\n

a)\u00a030<\/p>\n

b)\u00a022<\/p>\n

c)\u00a029<\/p>\n

d)\u00a0None of these<\/p>\n

Question 10:\u00a0<\/b>Find the difference of mean and median of the following data set {2,3,4,5,5,8,8}<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a03<\/p>\n

d)\u00a00<\/p>\n

Question 11:\u00a0<\/b>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?<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a03<\/p>\n

d)\u00a0Can\u2019t be determined<\/p>\n

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Question 12:\u00a0<\/b>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<\/p>\n

a)\u00a026<\/p>\n

b)\u00a028<\/p>\n

c)\u00a030<\/p>\n

d)\u00a032<\/p>\n

e)\u00a034<\/p>\n

Question 13:\u00a0<\/b>What is the difference between the mean and median of set S = {2, 4, 6, 7, 7, 13, 18, 92}?<\/p>\n

a)\u00a010.125<\/p>\n

b)\u00a011.125<\/p>\n

c)\u00a011.625<\/p>\n

d)\u00a014<\/p>\n

e)\u00a010.875<\/p>\n

Question 14:\u00a0<\/b>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:<\/p>\n

a)\u00a050<\/p>\n

b)\u00a051<\/p>\n

c)\u00a057<\/p>\n

d)\u00a056<\/p>\n

Question 15:\u00a0<\/b>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?<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

d)\u00a0None of the above<\/p>\n

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Question 16:\u00a0<\/b>What is a if the mean of the set S={4, 10, a, 12, 13, 21} is same as the mode?<\/p>\n

a)\u00a010<\/p>\n

b)\u00a010.75<\/p>\n

c)\u00a012<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

Question 17:\u00a0<\/b>What is the difference between the median and mode of S={1, 2, 4, 4, 8, 14, 32, 64}?<\/p>\n

a)\u00a00<\/p>\n

b)\u00a02<\/p>\n

c)\u00a04<\/p>\n

d)\u00a0Either 0 or 4<\/p>\n

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Question 18:\u00a0<\/b>What is the difference in the median and mode of the list { 6,2,3,4,3,5,2,7,4,2,6}<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a03<\/p>\n

d)\u00a02.5<\/p>\n

Question 19:\u00a0<\/b>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?<\/p>\n

a)\u00a0263<\/p>\n

b)\u00a0258.5<\/p>\n

c)\u00a0345<\/p>\n

d)\u00a0278.5<\/p>\n

Question 20:\u00a0<\/b>What is the difference between the mean and median of set S={2, 4, 6, 7, 7, 13, 18, 92}?<\/p>\n

a)\u00a010.125<\/p>\n

b)\u00a011.125<\/p>\n

c)\u00a011.625<\/p>\n

d)\u00a014<\/p>\n\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Mode = 3(Median)-2(Mean)<\/p>\n

Mode+2(Mean) = 3(Median)<\/p>\n

So the answer is option C.<\/p>\n

2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Mode = 3(Median)-2(Mean)<\/p>\n

Mode = 3(5)-2(4) = 15 – 8 = 7<\/p>\n

So the answer is option B.<\/p>\n

3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Mode = 3(Median)-2(Mean)<\/p>\n

7 = 3(5)-2(mode)<\/p>\n

7-15 = -2(mode)<\/p>\n

-8 = -2(mode)<\/p>\n

mode = 4<\/p>\n

So the answer is option D.<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Mode is the number which occurred frequently = 2<\/p>\n

So the answer is option A.<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Mode is the most frequently occurred number = 3<\/p>\n

So the answer is option B.<\/p>\n

6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$3(median)-2(mean) = mode$<\/p>\n

$3(6)-2(4) = mode$<\/p>\n

$18-8 = mode$<\/p>\n

$10 = mode$<\/p>\n

So the answer is option B.<\/p>\n

7)\u00a0Answer:\u00a096<\/b><\/p>\n

If the average of 9 numbers is 87, then the sum of these 9 distinct numbers will be 9 X 87 = 783
\nLet the numbers be a1, a2, a3, \u2026..a9 where a9> a8> \u2026a1.
\nSo, a1+a2+a3+a4+a5 = 78 X 5 = 390
\nSmallest value of a5 can be 80 when a1, a2, a3, a4, and a5 are 76, 77, 78, 79, 80.
\nThis means that a6 > 80.
\nNow, the sum of the rest of the four numbers is 738-390 = 393
\nFor a6 to be min, a7, a8, a9 must be max.
\n=> a6 + 98 + 99 + 100 = 393 or a6 = 96.
\nThus, 96 is the correct answer.<\/p>\n\n

8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$f(x)= x^4-8x^3+22x^2-24x+4$
\n$f'(x)=4x^3-24x^2+44x-24$
\nand $f”(x)=12x^2-48x+44$
\nf\u2019(x)=0
\n=> $4x^3-24x^2+44x-24=0$ or $(x-1)(x-2)(x-3)=0$
\nThe critical points are x=1,2,3
\nPutting these points in f\u2019\u2019(x) we get f(1)=+ve, f(2)=-ve, f(3)=+ve
\nSo, 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.<\/p>\n

<\/p>\n

Thus, the lower value between f(1) and f(3) will be the minimum value of f(x)
\nBoth f(1)=f(3)=-5
\nThus, minimum value is -5.<\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

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

10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Median is the middle term which is 5.
\nMean=(2+3+4+5+5+8+8)\/7=5
\nHence difference=0<\/p>\n

11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let the marks of the students in ascending order be a,b,c,d,e
\na+b+c+d+e=5*5=25
\nMedian=5
\nc=5
\nMode=8
\nd and e each is equal to 8 because if there are more than 2 8’s the sum becomes more than 25
\na+b+5+8+8=25 a+b=4
\nThe answers can be a=1 and b=3 or a=0 and b=4
\nHence, the answer is can\u2019t be determined. a=b=2 cannot be possible because there is only one mode<\/p>\n

12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

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<\/p>\n

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<\/p>\n

Here we need to maximize f, so we need to minimize a, b, c, d and e.<\/p>\n

The minimum value a and b can take is 1, the least positive integer.<\/p>\n

To minimize e, e should be equal to d+1. It cannot be equal to d as the mode is less than 18.<\/p>\n

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<\/p>\n

Hence, a = 1, b = 1, c = 17, d = 19, e = 20. Therefore, f = 90 -(1+1+17+19+20) = 32.<\/p>\n

13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Sum of the numbers = 2+4+6+7+7+13+18+92 = 149<\/p>\n

So, mean = 149\/8 =\u00a018.625<\/p>\n

Median = 7 (frequency of occurrence = 2)<\/p>\n

Hence, difference = 18.625 – 7 = 11.625<\/p>\n

14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

6a+8b+20c = 504 and b = a+2c<\/p>\n

The only integral solution to the above equation is when a = 18 b = 32 and c = 7.<\/p>\n

Hence, the total number of students in the class were 57<\/p>\n

15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

It is given mean,median and mode are in A.P. Hence we have mean + mode = 2*median
\nThe mode of the set of numbers is 3. And the mean of the numbers is (x+31)\/7.
\nIf x<3, the median is 3. But, as the arithmetic progression is non constant, no such values are possible.
\nIf 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.
\nIf 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.
\nSo, total number of valid solutions is 2<\/p>\n

16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let m be the mode and M the mean. Hence, M=(60+a)\/6=10+a\/6.
\nFor 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.
\nAs M is 10+a\/6 it cannot be 4 or 10. If a=12,13,21 => M = 12, 12.167, 13.5.
\nHence only for a=12 is the mean and mode same. Hence a=12.<\/p>\n

17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Mode=4, Median = (4+8)\/2=6. Hence difference=2<\/p>\n

18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The list when ordered in ascending order is {2,2,2,3,3,4,4,5,6,6,7}.
\nSo, the median is 4 and the mode is 2. Hence, difference is 2<\/p>\n\n

19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

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

20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

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
\nThe median of the set is the middle number of the set when it is placed in ascending order. In this case, it is 7.
\nHence, the difference is 11.625<\/p>\n\n

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We hope this Elementary Statistics questions for RRB NTPC Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

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. Take a free mock test for RRB NTPC Download RRB NTPC Previous Papers PDF Question 1:\u00a0What is the relation […]<\/p>\n","protected":false},"author":41,"featured_media":26787,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1603,1601],"tags":[489,491,1769,1619],"class_list":{"0":"post-26785","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-railways","10":"category-rrb-ntpc","11":"category-rrb-ntpc-je","12":"tag-railway-exam","13":"tag-rrb","14":"tag-rrb-mock-tedt","15":"tag-rrb-ntpc-2019"},"better_featured_image":{"id":26787,"alt_text":"RRB NTPC Elementary Statistics PDF","caption":"RRB NTPC Elementary Statistics PDF","description":"RRB NTPC Elementary Statistics 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