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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 701. |
The population of 50 villages in a state is given below: `{:("Population","Number of villages"),(" "6000," "8),(" "7000," "10),(" "9000," "12),(" "10000," "5),(" "11000," "7),(" "13000," "6),(" "15000," "2),(" ""Total"," "50):}` Find the mean population of the villages of the state. |
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Answer» The mean `bar(x)` is given by `bar(x)=((6000 xx 8)+(7000 xx 10)+(9000 xx 12)+(10000 xx 5)+(11000 xx 7)+(13000 xx 6)+(15000 xx 2))/(8+10+12+5+7+6+2)` `=(461000)/(50)` `bar(x) = 9220`. That is, the mean population of the villages is 9220. |
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| 702. |
The mean and variance of 7 observations are 8 and 16,respectively. If five of the observations are 2, 4, 10, 12, 14. Find theremaining two observations. |
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Answer» Let remaining tw observations are `x_(1)` and `x_(2)`. Here `n=7` Give that `barx=8implies(sumx_(i))/n=8` `implies x_(1)+x_(2)+2+4+10+12+14=8xx7` `implies x_(1)+x_(2)=14`…………1 and variance `=16` `implies (sumx_(i)^(2))/n-((sumx)/n)^(2)=16implies(sumx_(i)^(2))/7=16+(8)^(2)=80` `implies sumx_(i)^(2)=560` `implies x_(1)^(2)+x_(2)^(2)+2^(2)+4^(2)+10^(2)+12^(2)+14^(2)=560` `=x_(1)^(2)+14-x_(1)^(2)+4+16+100144+196=560` `implies x_(1)^(2)+196-28x_(1)+x_(1)^(2)+460-560=0` `implies 2x_(1)^(2)-28x_(1)+96=0` `=x_(1)^(2)-14x_(1)+48=0` `implies(x_(1)-6)(x_(1)-8)=0` `implies x_(1)-6=0` or `x_(1)-8=0` `implies x_(1)=6` or `x_(1)=8` If `x_(1)=6` then `x_(2)=14-6=8` `x_(1)=8` then `x_(2)=14-8=6` `:.` Remaining two observations `=6,8`. |
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| 703. |
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2,4,10,12,14, find the remaining two observations. |
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Answer» Let the remaining two observations be x and y. Then, `"mean"=8rArr(2+4+10+12+14+x+y)/(7)=8` `rArr" "42+x+y=56` `rArr" "x+y=14." ...(i)"` Also, variance =16 `rArr" "(1)/(7)(2^(2)+4^(2)+10^(2)+12^(2)+14^(2)+x^(2)+y^(2))-8^(2)=16" "[because sigma^(2)=(Sigmax_(i)^(2))/(n)-(barx)^(2)]` `rArr" "(1)/(7)(460+x^(2)+y^(2))=80` `rArr" "460+x^(2)+y^(2)=560` `rArr" "x^(2)+y^(2)=100." ...(ii)"` `"Now, "(x+y)^(2)+(x-y)^(2)=2(x^(2)+y^(2))` `rArr" "(x-y)^(2)=2(x^(2)+y^(2))-(x+y)^(2)` `rArr" "(x-y)^(2)=(2xx100)-(14)^(2)=(200-196)=4` `rArr" "x-y=pm2.` `"Now, "x+y=14,x-y=2rArrx=8,y=6,` `x+y =14, x-y=-2rArrx=6,y=8.` Hence, the remainig two observations are 6 and 8. |
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| 704. |
The correct relationship among A.M., G.M. and H.M.is: (a) A.M. < G.M. < H.M. (b) G.M. > A.M. > H.M. (c) H.M. > G.M. > A.M. (d) A.M. > G.M. > H.M. |
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Answer» (d) A.M. > G.M. > H.M. |
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| 705. |
Harmonic mean is the reciprocal of: (a) Median of the values. (b) Geometric mean of the values. (c) Arithmetic mean of the reciprocal of the values. (d) Quartiles of the values. |
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Answer» (c) Arithmetic mean of the reciprocal of the values. |
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| 706. |
When an observation in the data is zero, then its geometric mean is: (a) Negative (b) Positive (c) Zero (d) Cannot be calculated |
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Answer» Answer is (c) Zero |
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| 707. |
Which of the following is a positional measure? (a) Range (b) Mode (c) Mean deviation (d) Percentiles |
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Answer» (d) Percentiles |
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| 708. |
The median of 24, 20, 32, 18, 28, 16, 25 isA) 18 B) 16 C) 24 D) 32 |
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Answer» Correct option is: C) 24 The ascending order of given observations is as follows : 16, 18, 20, 24, 25, 28, 32 Total observations are n = 7 which is odd \(\therefore\) \(\frac {n+1}2\) = \(\frac {7+1}2 = \frac 82 = 4\) Hence, 4th observation in ascending order of given observations is median of the given data. \(\therefore\) 4th observation = 24 is median of the given data. Correct option is: C) 24 |
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| 709. |
Class interval of 1 – 10 is A) 8 B) 12 C) 11 D) 10 |
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Answer» Correct option is: D) 10 In inclusive class 1-10, upper limit 10 is also included. \(\therefore\) Class interval = Upper limit -Lower limit + 1 = 10 -1 + 1 = 10 Alternative method : Inclusive class is 1-10. For finding class interval, we have to convert inclusive class into exclusive class. For that we have to subtract 0.5 from lower limit of class and add 0.5 to upper limit of class \(\therefore\) Exclusive form of class is 0.5 -10.5. \(\therefore\) Class interval of inclusive class 1-10 = Class interval of exclusive class 0.5 - 10.5 = True upper limit - True lower limit = 10.5 - 0.5 = 10 Correct option is: D) 10 |
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| 710. |
When calculating the average growth of the economy, the correct mean to use is? (a) Weighted mean (b) Arithmetic mean (c) Geometric mean (d) Harmonic mean |
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Answer» (c) Geometric mean |
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| 711. |
Length of the class 11-20 is …………A) 9 B) 10 C) 11 D) 20 |
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Answer» Correct option is: A) 9 Length of the class 11-20 = Upper limit of the class - Lower limit of the class = 20-11 = 9 Correct option is: A) 9 |
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| 712. |
For the terms, x + 1, x + 2, x – 1, x + 3 and x – 2(x ∈ N), if the median of the data is 12, then x = ………A) 9 B) 10 C) 11 D) 13 |
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Answer» Correct option is: C) 11 Ascending order of given observation is as follows x-2, x-1, x+1, x+2, x+3 Total 5 observation \(\therefore\) n = 5 which is odd Now, \(\frac {n+1}2 = \frac {5+1}{2} = \frac 62 =3^{rd}\) Hence, \(3^{rd}\) observation is median of the data. Hence, x+1 is median of the data. \(\therefore\) x+1 = 12 (\(\because\) 12 is median of the given data) \(\Rightarrow\) x = 12 -1 = 11 Correct option is: C) 11 |
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| 713. |
If the difference of mode and median of a data is 24, then the difference of medina and mean is A. 12 B. 24 C. 8 D. 36 |
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Answer» Difference of mode and median, mode - median = 24 …[1] We know that empirical relation between mean, median and mode is Mode = 3 Median - 2 Mean ⇒ 3 Mode - 2 Mode = 3 Median - 2 Mean ⇒ 3 Mode - 3 Median = 2 Mode - 2 Mean ⇒ 3(Mode - Median) = 2(Mode - Mean) From [1] we have 3(24) = 2(Mode - Mean) ⇒ Mode - Mean = 36 …[2] on substracting [1] from [2] Mode - Mean - (Mode - Median) = 36 - 24 Mode - Mean - Mode + Median = 8 Median - Mode = 8 Hence, difference between median and mode is 8. |
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| 714. |
If the mean and median of a set of numbers are 8.9 and 9 respectively, then the mode will be (A) 7.2 (B) 8.2 (C) 9.2 (D) 10.2 |
| Answer» The correct answer is (C) 9.2 | |
| 715. |
If the arithmetic mean of the following distribution is 8.2, then a = `{:(x_(i) " :",1,3,5,9,11,13),(f_(i)" :",3,2,7,a,4,8):}`A. 5B. 6C. 9D. None of these |
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Answer» Correct Answer - B (i) `(sum fx)/(N) = 8.2`. (ii) Find `sum fx "and" sum f`. (iii) Mean `= (sum fx)/(sum f)` |
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| 716. |
If the arithmetic mean of 7, 8, x, 11, 14 is x, then x = A. 9 B. 9.5 C. 10 D. 10.5 |
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Answer» Terms are 7, 8, x, 11, 14 No of terms = 5 We know that Mean \(=\frac{Sum\,of\,all\,observations}{No\,of\,observations}\) \(\Rightarrow x = \frac{7+8+x+11+14}{5}\) ⇒ 5x = x + 40 ⇒ 4x = 40 ⇒ x = 10 Hence, correct option is (C) |
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| 717. |
If the arithmetic mean of 7 , 8 , x , 11 , 14 is x , then x is `"__________"`.A. 9B. 9.5C. 10D. 10.5 |
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Answer» Correct Answer - c Arithmatic mean = `("Sum of the quantities")/("Number of the quantities")` |
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| 718. |
If the arithmetic mean of 5, 7, 9, x is 9 then the value of x is (A) 11 (B) 15 (C) 18 (D) 16 |
| Answer» (B) The value of x is 15. | |
| 719. |
There are 10 cards numbered from 1 to 10 in a box. If a card is drawn randomly, then find the probability of getting an even numbered card.A. `1/10`B. `1/5`C. `2/5`D. `1/2` |
| Answer» Correct Answer - D | |
| 720. |
Find the maximum difference between sales of scooters and that of cars, in ant year, in the given period.A. 1500B. 1700C. 1800D. 2000 |
| Answer» Correct Answer - C | |
| 721. |
Total number of vehicles (scooters and cars) sold in the year 2005 and 2006 is ______.A. 26100B. 28500C. 25100D. 27500 |
| Answer» Correct Answer - A | |
| 722. |
Find the ratio between the total number of vehicles sold (scooters and cars) in the year 2006 and that in the year 2008.A. `41 : 46`B. `69 : 91`C. `147 : 182`D. `46 : 49` |
| Answer» Correct Answer - B | |
| 723. |
Find the total number of scooters sold in the four years.A. 26000B. 27000C. 31000D. 32000 |
| Answer» Correct Answer - D | |
| 724. |
If P(A) denotes the probability of an event A, then (A) P(A) < 0 (B) P(A) > 1 (C) 0 ≤ P(A) ≤ 1 (D) –1 ≤ P(A) ≤ 1 |
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Answer» (C) 0 ≤ P(A) ≤ 1 Since, probability of an event always lies between 0 and 1. |
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| 725. |
If an event cannot occur, then its probability is(A)1 (B) ¾ (C) ½ (D) 0 |
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Answer» (D) 0 Explanation: The event which cannot occur is said to be impossible event. The probability of impossible event = zero. Hence, option (D) is correct |
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| 726. |
Which of the following cannot be the probability of an event?(A)1/3 (B) 0.1 (C) 3% (D)17/16 |
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Answer» (D)17/16 Explanation: Probability of an event always lies between 0 and 1. Probability of any event cannot be more than 1 or negative as (17/16) > 1 Hence, option (D) is correct |
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| 727. |
If the probability of an event is P, then the probability of its complementary event will be A. P - 1 B. P C. 1 - P D. 1 - 1/P |
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Answer» C. 1 - P Since, probability of an event + probability of its complementary event = 1 So, probability of its complementary event = (1 – Probability of an event) = 1 – P. |
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| 728. |
An event is very unlikely to happen. Its probability is closest to A. 0.0001 B. 0.001 C. 0.01 D. 0.1 |
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Answer» A. 0.0001 The probability of an event which is very unlikely to happen is closest to zero and from the given options 0.0001 is closest to zero. |
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| 729. |
The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated belowThe number of athletes who completed the race in less than 14.6 seconds is :(A) 11 (B) 71 (C) 82 (D) 130 |
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Answer» (C) 82 Explanation: The number of athletes who completed the race in less than 14.6 second= 2 + 4 + 5 + 71 = 82 Hence, option (C) is correct |
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| 730. |
Given below is the data showing heights of 50 students in a class. Find its median. `{:("Height(in cm)",162,164,166,167,168,170,173,175,177,180),("Number of Students",6,4,5,12,8,3,7,2,2,1):}` |
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Answer» To find median, we prepare less than cumulative frequency table as given below: `{:("Height(in cm)","Number of Students","Cumulative Frequency(f)"),(" "162," "6," "6),(" "164," "4," "10),(" "166," "5," "15),(" "167," "12," "27),(" "168," "8," "35),(" "170," "3," "38),(" "173," "7," "45),(" "175," "2," "47),(" "177," "2," "49),(" "180," "1," "50):}` Here N=50, which is even. `therefore "Median" (N)/(2) "value" (50)/(2)` or 25th observation. From the column of cumulative frequency, the value of 25th observation is 167. `therefore` Median = 167 cm. |
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| 731. |
The sum and the sum of squares of length x (in cm) and weight y (in g) of 50 plant products are given below : `Sigma_(i=1)^(50)x_(i)=212, Sigma_(i=1)^(50) x_(i)^(2)=902.8, Sigma_(i=1)^(50) y_(i)=261 and Sigma_(i=1)^(50)y_(i)^(2)=1457.6.` Which is more variable, the length or weight ? |
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Answer» Correct Answer - Weight var(x)=(1)/(50)cdotoverset(50)underset(i=1)Sigmax_(i)^(2)-((1)/(50)cdotoverset(50)underset(i=1)Sigmax_(i))^(2)=((902.8)/(50))-((212)/(50))^(2)=(451.4)/(25)-((106)/(25))^(2)=((451.4)/(25)-(11236)/(625))=((11285-11236)/(625))=((49)/(625)=0.078.` var(y)=(1)/(50)cdotoverset(50)underset(i=1)Sigmay_(i)^(2)-((1)/(50)cdotoverset(50)underset(i=1)Sigmay_(i))^(2)={(1457.6)/(50)-((261)/(50))^(2)}={129.152-(5.22)^(2)}=29.152-27.2484` `=1.9036.` `therefore" "var(x)ltvar (y)` |
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| 732. |
Suppose a population A has 100 observations `101,102,................,200` and another population B has 100 observations `151,152,........250.` If `V_A and V_B` represent the variances of the two populations respectively,then `V_A/V_B` is (a) `1` (b) `9/4` (c) `4/9` (d) `2/3`A. 1B. `9//4`C. `4//9`D. `2//3` |
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Answer» Correct Answer - A `sigma^(2)=(sum d_(i)^(2))/(n)` (Here deviations are taken from the mean.) Since both A and B have 100 consecutive integers, they have the same standard deviation and hence variance. Therefore, `(V_(A))/(V_(B))=1 ` (As `sum d_(i)^(2)` is same in both the cases.) |
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| 733. |
Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why? |
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Answer» Apoorv throw two dice at once. Hence, the total number of outcomes = 36 Number of outcomes for getting product 36 = 1(6×6) ∴ Probability for Apoorv = 1/36 Peehu throws one die, Hence, the total number of outcomes = 6 Number of outcomes for getting square = 36 ∴ Probability for Peehu = 6/36 = 1/6 Therefore, Peehu has a better chance of getting the number 36. |
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| 734. |
When we toss a coin, there are two possible outcomes - head or tail. Therefore, the probability of each outcome is 1/2. Justify your answer. |
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Answer» Yes, probability of each outcome is 1/2 because head and tail both are equally likely events. |
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| 735. |
If I toss a coin 3 times and get head each time, should I expect a tail to have a higher change in the 4th toss? Given reason in support of your answer. |
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Answer» No, let we toss a coin, then we get head or tail, both are equally likely events. i.e., probability of each event is 1/2 .So, no question of expecting a tail to have a higher change in 4th toss. |
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| 736. |
I toss three coins together. The possible outcomes are no heads, 1 head, 2 head and 3 heads. So, I say that probability of no heads is 1/4. What is wrong with this conclusion? |
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Answer» I toss three coins together [given] So, total number of outcomes = 23 = 8 and possible outcomes are(HHH),(HTT),(THT),(TTH),(HHT),(THH),(HTH) and (TTT). Now, probability of getting no head = 1/8 Hence, the given conclusion is wrong because the probability of no head is 1/8 and 1/4. |
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| 737. |
If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1? Given reasons. |
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Answer» No, if let we toss a coin, then we get head or tail, both are equally likely events. So, probability is 1/4 if we toss a coin 6 times, then probability will be same in each case. So, the probability of getting a head is not 1. |
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| 738. |
Two coins are tossed 1000 times and the outcomes are recorded as below :Based on this information, the probability for at most one head is(A) 1/5(B) 1/4(C) 4/5(D) 3/4 |
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Answer» (C) 4/5 The total number of coins tossed, n(S) = 1000 Number of outcomes in which atmost one head, n(E) = 550 + 250 = 800 =n(E)/n(S) = 800/1000 = 4/5 Hence, the probability for atmost one head is 4/5 . |
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| 739. |
The probability that a non - leap your selected at random will contains 53 Sunday is(A) 1/7(B) 2/7(C) 3/7(D) 5/7 |
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Answer» (A) 1/7 A non - leap year has 365 days and therefore 52 weeks and 1 day. This 1 day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday. Thus, out of 7 possibilities, 1 favorable event is the event that the one day is Sunday. ∴ Required probability = 1/7 |
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| 740. |
The value of π up to50 decimal places is given below:3.14159265358979323846264338327950288419716939937510(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.(ii) What are the most and the least frequently occurring digits? |
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Answer» (i) By observation of the digits after decimal point, the required table can be constructed as follows.
(ii) It can be observed from the above table that the least frequency is 2 of digit 0, and the maximum frequency is 8 of digit 3 and 9. Therefore, the most frequently occurring digits are 3 and 9 and the least frequently occurring digit is 0. |
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| 741. |
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:0 1 2 2 1 2 3 1 3 01 3 1 1 2 2 0 1 2 13 0 0 1 1 2 3 2 2 0Prepare a frequency distribution table for the data given above. |
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Answer» By observing the data given above, the required frequency distribution table can be constructed as follows.
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| 742. |
Three coins are tossed 30 times. Each time the number of heads occurring was noted down as follows:0, 1, 2, 2, 1, 2, 3, 1, 3, 0, 1, 3, 1, 1, 2, 2, 0, 1, 2, 1, 0, 3, 0, 2, 1, 1, 1, 3, 2, 0, 2.Prepare a frequency distribution table. |
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Answer» Frequency distribution table
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| 743. |
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:0.03 0.08 0.08 0.09 0.04 0.170.16 0.05 0.02 0.06 0.18 0.200.11 0.08 0.12 0.13 0.22 0.070.08 0.01 0.10 0.06 0.09 0.180.11 0.07 0.05 0.07 0.01 0.04(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08, and so on.(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million? |
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Answer» Taking class intervals as 0.00, −0.04, 0.04, −0.08, and so on, a grouped frequency table can be constructed as follows.
The number of days for which the concentration of SO2 is more than 0.11 is the number of days for which the concentration is in between 0.12 − 0.16, 0.16 − 0.20, 0.20 − 0.24. Required number of days = 2 + 4 + 2 = 8 Therefore, for 8 days, the concentration of SO2 is more than 0.11 ppm. |
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| 744. |
Mean of 2.8, 5/2, 5.2, 2/5 is ……………….. A) 0.4 B) 2.5 C) 2.65 D) 5.2 |
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Answer» Correct option is (C) 2.65 Ascending order of observation is \(\frac{2}{5},\frac52,2.8,5.2\) \(\therefore\) Mean \(=\frac12\,(\frac52+2.8)\) \(=\frac{2.5+2.8}2\) \(=\frac{5.3}2\) = 2.65 Correct option is C) 2.65 |
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| 745. |
Mean of 100 items is 49. It was discovered that three items which should have been `60, 70,80` were wrongly read as `40, 20, 50` respectively. The correct mean is. (a) `48` (b) `82 1/2` (c) `50` (d) `80`A. 48B. 82.5C. 50D. 80 |
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Answer» Correct Answer - C Sum of 100 items `=49xx100=4900` Sum of items added=60+70+80=210 New sum =4900+210-110=5000 `therefore " Correct mean" =(5000)/(100)=50` |
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| 746. |
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.53.5 2.3 3.2 3.4 3.8 3.2 4.6 3.72.5 4.4 3.4 3.3 2.9 3.0 4.3 2.83.5 3.2 3.9 3.2 3.2 3.1 3.7 3.44.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the intervals 2 − 2.5. |
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Answer» A grouped frequency table of class size 0.5 has to be constructed, starting from class interval 2 − 2.5. Therefore, the class intervals will be 2 − 2.5, 2.5 − 3, 3 − 3.5… By observing the data given above, the required grouped frequency distribution table can be constructed as follows.
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| 747. |
If for a sample of size 60, we have the following information `sumxi2=18000a n dsumxi=960`, then the variance is6.63 (b) 16 (c) 22 (d) 44A. 6.63B. 16C. 22D. 44 |
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Answer» Correct Answer - D Variance `=(sum x_(i)^(2))/(n)-((sum x_(i))/(n))^(2)` `=(18000)/(60)-((960)/(60))^(2)=300-256=44` |
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| 748. |
The standard deviation of variate `x_i` is σ. Then standard deviation of the variate `(ax_i+b)/c` where `a, b, c ` are constants is- (a) `(a/c)sigma` (b) `|a/c| sigma` (c) (a^2/c^2)sigma` (d) Non of theseA. `((a)/(c ))sigma`B. `|(a)/(c )|sigma`C. `((a^(2))/(c^(2)))sigma`D. None of these |
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Answer» Correct Answer - B Let `y=(alpha x+b)/(c ), i.e., y=(a)/(c )x+(b)/(c )` i.e., `y=(a)/(c )x+(b)/(c )` `therefore overline (y)=Aoverline(x)+B` `therefore y-overline(y)=A(x-overline(x))` `implies (y-overline(y))^(2)=A^(2)(x-overline(x))^(2)` `implies sum (y-overline(y))^(2)=A^(2)sum(x-overline(x))^(2)` `implies n. sigma_(y)^(2)=A^(2). n sigma_(x)^(2)` `implies sigma_(y)^(2)=A^(2)sigma_(x)^(2)` `implies sigma_(y)=|A|sigma_(x)` `implies sigma_(y)=|(a)/(c )| sigma_(x)` Thus, new SD `=|(a)/(c )|sigma` |
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| 749. |
If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of thefollowing is true ?(1) `3a^2-26 a+55=0`(2) `3a^2-32 a+84=0`(3) `3a^2-34 a+91=0`(4) `3a^2-23 a+44=0`A. `3a^(2)-32a+84=0`B. `3a^(2)-34a+91=0`C. `3a^(2)-23a+44=0`D. `3a^(2)-26a+55=0` |
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Answer» Correct Answer - A Standard deviation of numbers 2,3,a and 11 is 3.5 `therefore (3.5)^(2)=(sum x_(i)^(2))/(4)-(overline(x))^(2)` `implies (3.4)^(2)=(4+9+a^(2)+121)/(4)-((2+3+a+11)/(4))^(2)` On solving, we get `3a^(2)-32a+84=0` |
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| 750. |
Cosider the number 1,2,3,4,5,6,7,8,9 and 10.If 1 is added to each number the variance of the number so obtained isA. 6.5B. 2.87C. 3.87D. 8.25 |
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Answer» Correct Answer - D Given number are 1,2,3,4,5,6,7,8,9, 10 and 11 `therefore " " Sigmax_(i)=2+3+4+……+11` `=10/2[2xx2+9xx1]=5[4+9]=65` `Sigma x_(i)^(2)=2^(2)+3^(2)+4^(2)+5^(2)+….+11^(2)` `=(1^(2)+2^(2)+3^(3)+….+11^(2))-(1^(2)` `=(11xx12xx23-6)/(6)`=505 `therefore s^(2)=(Sigma_(i)^(2))/n-((Sigmax_(i))/n)^(2)=505/10-((65)/10)^(2)` ` =505-(65/10)^2` ` 505-(6.5)^2` `50.5-42.25` =8.25 |
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