(a) Both Assertion (A) and reason (R) are true and Reason (R) is a correct explanation of Assertion (A). (b) Both Assertion (A) and reason (R) are true and Reason (R) is not a correct explanation of...
(a) Both Assertion (A) and reason (R) are true and Reason (R) is a correct explanation of Assertion (A). (b) Both Assertion (A) and reason (R) are true and Reason (R) is not a correct explanation of...
(a) 13 (b) 14 (c) 15 (d) 16 Answer: (c) 15 Sol: Median of 6 numbers is the average of 3rd and 4th term. ∴ 13 = (????−1)+(????−3) 2 ⇒ 26 = 2x – 4 ⇒ 2x = 30 ⇒ x = 15 Thus, x is equal to...
(a) 0 (b) 1 (c) 10 (d) 19 Answer: (d) 19 Sol: It is given that mean of 20 numbers is zero. i.e., average of 20 numbers is zero. i.e., sum of 20 numbers is zero. Thus, at most, there can be 19...
(a) 7 (b) 9 (c) 11 (d) 13 Answer: (b) 9 Sol: First 8 prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19. Median of 8 numbers is average of 4th and 5th terms.
Answer: (c) 13 Sol: Converting the given data into a frequency table, we get: Hence, the number of families having an income range of Rs. 20,000 – Rs. 25,000 is 13. The correct option is...
(a) 27.5 (b) 24.5 (c) 28.4 (d) 25.8 Answer: (b) 24.5 Sol: Mode = (3 × median) – (2 × mean) ⇒ (2 × mean) = (3 × median) – mode ⇒ (2 × mean) = 3 × 26 – 29 ⇒ (2 × mean) = 49 ⇒ Mean = 49 2 ∴ Mean = 24.5...
(a) 22 (b) 23.5 (c) 24 (d) 24.5 Answer: (c) 24 Sol: Mode = (3 × median) – (2 × mean) ⇒ (3 × median) = (mode + 2 mean) ⇒ (3 × median) = 16 + 56 ⇒ (3 × median) = 72 ⇒ Median = 72 3 ∴...
(a) 7.2 (b) 8.2 (c) 9.2 (d) 10.2 Answer: (c) 9.2 Sol: It is given that the mean and median are 8.9 and 9, respectively, ∴ Mode = (3 × Median) – (2 × Mean) ⇒ Mode = (3 × 9) – (2 × 8.9) = 27 – 17.8 =...
(a) 5.5 (b) 15.5 (c) 20.5 (d) 36.0 Answer: (c) 20.5 Sol: The x- coordinate represents the median of the given data. Thus, median of the given data is 20.5.
(a) mode=(3 * mean) – (2 * median) (b) mode=(3 *median) – (2 *mean) (c) median=(3 * mean) – (2 * mode) (d) mean=(3 * median) – (2 *mode) Answer: (b) mode=(3 * median) – (2 *mean) Sol: mode=(3 *...
(a) evenly distributed over the classes (b) centred at the class marks of the classes (c) centred at the lower limits of the classes (d) centred at the upper limits of the classes Answer: (b)...
(a) lower limits of the classes (b) upper limits of the classes (c) midpoints of the classes (d) none of these Answer: (c) midpoints of the classes Sol: The ???????? ′???? are the deviations from A...
(a) Mean (b) Median (c) Mode (d)None of these Answer: (b) Median Sol: The abscissa of the point of intersection of the ‘less than type’ and that of the ‘more than type’ cumulative...
(a) Mean (b) Median (c) Mode (d) all of these Answer: (b) Median Sol: The cumulative frequency table is useful in determining the median.
(a) a histogram (b) a frequency curve (c) a frequency polygon (d) ogives Answer: (d) ogives Sol: This because median of a frequency distribution is found graphically...
(a) a frequency curve (b) a frequency polygon (c) a histogram (d) an ogive Answer: (c) a histogram Sol: The mode of a frequency distribution can be obtained graphically from a histogram.
(a) Mean (b) Median (c) Mode (d) None of these Answer: (a) Mean Sol: Mean is influenced by extreme values.
(a) Mean (b) Median (c) Mode (d) None of these Answer: (a) Mean Sol: The mean cannot be determined graphically because the values cannot be summed.
(a) Mean (b) Mode (c) Median (d) Standard Deviation Answer: (d) Standard Deviation Sol: The standard deviation is a measure of dispersion. It is the action or process of distributing...
Sol: Since, 8 is less than 30 and 32 is more than 30, so the middle value remains unchanged Thus, the median of 21 observations taken together is 30.
Sol: Upper class boundary = Lowest class boundary + width × number of classes = 8.1 + 2.5×12 = 8.1 + 30 = 38.1 Thus, upper class boundary of the highest class is 38.1.
Solution: Given that $n=10, \bar{x}=45$ and $\sigma^{2}=16$ $\begin{array}{l} \bar{x}=45 \Rightarrow \frac{\Sigma x_{i}}{n}=45 \\ \Rightarrow \quad \frac{\Sigma x_{i}}{10}=45 \Rightarrow \Sigma...
and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.Solution: Provided, $n=100, \bar{x}=40$ and $\sigma=10$ $\begin{array}{ll} \therefore \quad & \frac{\Sigma x_{i}}{n}=40 \\ \Rightarrow \quad & \frac{\Sigma x_{i}}{100}=40 \\ \Rightarrow...
![\[\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|} \hline Ravi & 25 & 50 & 45 & 30 & 70 & 42 & 36 & 48 & 35 & 60 \\ \hline Hashina & 10 & 70 & 50 & 20 & 95 & 55 & 42 & 60 & 48 & 80 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-968cdee73496a40bbe3b252754ec6f8a_l3.png "Rendered by QuickLaTeX.com")
Solution: For Ravi $$\begin{tabular}{|c|c|c|} \hline$x_{i}$ & $d_{i}=x_{i}-45$ & $d_{i}^{2}$ \\ \hline 25 & $-20$ & 400 \\ \hline 50 & 5 & 25 \\ \hline 45 & 0 & 0 \\ \hline 30 & $-15$ & 225 \\...
terms of an A.P. whose first term is a and common difference is 
.Solution: Given that the first $n$ terms of an A.P. whose first term is a and common difference is d Now we have to find mean and standard deviation Given AP in tabular form is shown below,...
![\[\begin{tabular}{|l|l|} \hline Weight (in grams) & Frequency \\ \hline $200-201$ & 13 \\ \hline $201-202$ & 27 \\ \hline $202-203$ & 18 \\ \hline $203-204$ & 10 \\ \hline $204-205$ & 1 \\ \hline $205-206$ & 1 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c0bc6df090fc5305ccf0b694d1521306_l3.png "Rendered by QuickLaTeX.com")
Solution: The weights of coffee in 70 jars is given We now need to find the variance and standard deviation of the distribution Let's construct a table of the given data and append other columns...
![\[\begin{tabular}{|l|l|l|l|l|l|} \hline \text { Class interval } & 0-4 & 4-8 & 8-12 & 12-16 & 16-20 \\ \hline \text { Frequency } & 4 & 6 & 8 & 5 & 2 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-89e2ff7ca4a076d66a84b3171bda25fa_l3.png "Rendered by QuickLaTeX.com")
Solution: The frequency distribution is given We now need to find the mean deviation about the mean Let's construct a table of the given data and append other columns after calculations...
![\[\begin{tabular}{|l|l|l|l|l|} \hline$x$ & $1 \leq x<3$ & $3 \leq x<5$ & $5 \leq x<7$ & $7 \leq x<10$ \\ \hline$f$ & 6 & 4 & 5 & 1 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2d3871a8fdd83a0b06beb9abe864a5e8_l3.png "Rendered by QuickLaTeX.com")
Solution: The frequency distribution is given We now need to find the mean and variance Convert the ranges of $x$ to groups, we can re-write the given table as shown below,...
and the total number of item is 18 , find the mean and standard deviation.Solution: Provided that for a distribution $\sum(x-5)=3, \sum(x-5)^{2}=43$ and the total number of item is 18 Now we have to find the mean and standard deviation. As per the criteria given No. of...
Solution: It is given that mean and standard deviation of 100 items are 50 and 4, respectively We now need to find the sum of all items and the sum of the squares of the items As per the criteria...
Solution: Given that the mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7...
![\[\begin{tabular}{|l|l|l|l|l|l|l|} \hline Marks & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline Frequency & $x-2$ & $x$ & $x^{2}$ & $(x+1)^{2}$ & $2 x$ & $x+1$ \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d1c25739bda1e53fe838cb056af8bda4_l3.png "Rendered by QuickLaTeX.com")
is a positive integer. Determine the mean and standard deviation of the marks.Solution: It is given that there are 60 students in a class. Also the frequency distribution of the marks obtained by the students in a test is given. We now need to find the mean and the standard...
![\[\begin{tabular}{|l|l|l|l|l|l|l|} \hline $\mathbf{x}$ & $\mathbf{2}$ & $\mathbf{3}$ & $\mathbf{4}$ & $\mathbf{5}$ & $\mathbf{6}$ & $\mathbf{7}$ \\ \hline $\mathbf{f}$ & $\mathbf{4}$ & $\mathbf{9}$ & $\mathbf{1 6}$ & $\mathbf{1 4}$ & $\mathbf{1 1}$ & $\mathbf{6}$ \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0bbe3ac7f19d77a3ccb70fd0bb0e83a8_l3.png "Rendered by QuickLaTeX.com")
Solution: Provided: frequency distribution table We now need to find the standard deviation Let's construct a table of the data given and append other columns after calculations...
![\[\begin{tabular}{|l|l|l|l|l|l|l|} \hline$x$ & $A$ & $2 A$ & $3 A$ & $4 A$ & $5 A$ & $6 A$ \\ \hline$F$ & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8ea925103451b0f60ef84a2f59f3d909_l3.png "Rendered by QuickLaTeX.com")
is a positive integer, has a variance of 
Determine the value of .Solution: In the given frequency distribution table, where variance $=160$ We now need to find the value of $A$, where $A$ is a positive number We need to construct a table of the data given...
Solution: Provided: Two sets each of 20 observations, have the same standard derivation 5 . The first set has a mean 17 and the second a mean $22 .$ We now need to show that the standard deviation...
observations are 
and 
, respectively while the mean and standard deviation of another set of 
observations are 
and 
, respectively. Show that the standard deviation of the combined set of 
observations is given by 
Solution: The variance $\sigma^{2}$ of the combined series is given by...
, mean 
seconds, standard deviation 
seconds. Further, another set of 15 observations 
, also in seconds, is now available and we have 
and 
Solution: Provided: No. of observations $=25$, mean $=18.2$ seconds, standard deviation = $3.25$ seconds. Another set of 15 observations $x_{1}, x_{2}, \ldots, x_{15}$, also in seconds, is...
Solution: It is given: set of first $n$ natural numbers We now need to find the standard deviation Provided first $n$ natural numbers, it can be written in table as shown below...
natural numbers when 
is an even number.Solution: It is given that set of first $n$ natural numbers when $\mathrm{n}$ is an even number. We now need to find the mean deviation about the mean It is known that the first $n$ natural numbers...
natural numbers when 
is an odd number.Solution: It is given that the set of the first $n$ natural numbers when $n$ is an odd number. We now need to find the mean deviation about the mean. It is known that first $n$ natural numbers are...
![\[\begin{tabular}{|l|l|l|l|l|l|} \hline Marks obtained & 10 & 11 & 12 & 14 & 15 \\ \hline No. of Students & 2 & 3 & 8 & 3 & 4 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9180182ab9d77161dc1a26513669b15b_l3.png "Rendered by QuickLaTeX.com")
Solution: Data distribution is given We now need to find the mean deviation about the median Construct a table of the given data and append other columns after calculations...
![\[\begin{tabular}{|l|l|l|l|l|l|} \hline Size & 20 & 21 & 22 & 23 & 24 \\ \hline Frequency & 6 & 4 & 5 & 1 & 4 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-67012859a1da859a3d725cf996f07cf6_l3.png "Rendered by QuickLaTeX.com")
Solution: Data distribution is given We now need to find the mean deviation about the mean of the distribution. Draw a table of the data given. $$\begin{tabular}{|l|l|l|} \hline Size...
(I) In this sentence 'and' is the associating word The part proclamations are as per the following (a) All normal numbers are genuine (b) All genuine numbers are not perplexing (ii) In this...
Weight(in kg)40-4545-5050-5555-6060-6565-7070-75Number of students2386632 Solution: Class Interval Frequency ...
Number of letters1-44-77-1010-1313-1616-19Number of surnames630401644 Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size...
Lifetime (in hours)Number of lamps1500-2000142000-2500562500-3000603000-3500863500-4000744000-4500624500-500048 Find the median lifetime of a lamp. Solution: Class...
Length (in mm)Number of leaves118-1263127-1355136-1449145-15312154-1625163-1714172-1802 Find the median length of leaves. solution: Since the information are not ceaseless diminish 0.5 in as far as...
workers of a factory:Daily income (in Rs):No of workers:100 – 12012120 – 14014140 – 1608160 – 180 6180 – 20010 Convert the above distribution to a ‘less than’ type...
shops are distributed as follows:Profit per shopNo of shops:0 – 50 1250 – 10018100 – 15027150 – 20020200 – 25017250 – 3006 Draw the frequency polygon for it. Solution: Statistics is the discipline...
Class-interval0 – 45 – 910 – 1415 – 1920 – 24No. of students261053...
students in an examination are given in the form of a frequency distribution table:MarksNo. of Students600 – 640 16640 – 68045680 – 720156720 – 760284760 – 800172800 – 84059840 – 88018 Solution: Statistics is the discipline that concerns the...
persons, who bought the shirt from a store, are as follows:Shirt size:3738394041424344Number of persons:1525394136171512 Solution: Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data....
(i) Class interval:0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80Frequency:5871228201010 Solution Statistics is the discipline that concerns the collection, organization, analysis,...
Age in years16 – 1818 – 2020 – 2222 – 2424 – 26Group A5078462823Group B5489402517 Solution Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and...
(i) $3,5,7,4,5,3,5,6,8,9,5,3,5,3,6,9,7,4$ (ii) $3,3,7,4,5,3,5,6,8,9,5,3,5,3,6,9,7,4$ (iii) $15,8,26,25,24,15,18,20,24,15,19,15$ Solution: Statistics is the...
students of class X are given below. Find the mode of the marks obtained by the students in science.Marks0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 9090 – 100Frequency35161213205411 Solution Statistics is the discipline that concerns the collection, organization, analysis,...
for giving distribution, mean is 
$x:$$3$$5$$7$$9$$11$$13$$f:$$6$$8$$15$$p$$8$$4$ Solution: $x$$f$$fx$$3$$6$$18$$5$$8$$40$$7$$15$$105$$9$$p$$9p$$11$$8$$88$$13$$4$$52$$N=41+p$$\sum{fx=303+9p}$ We know that, Mean $=\sum{fx/N=\left(...
for the given distribution whose mean is 
$x:$$5$$8$$10$$12$$p$$20$$25$$f:$$2$$5$$8$$22$$7$$4$$2$ Solution: $x$$f$$fx$$5$$2$$10$$8$$5$$40$$10$$8$$80$$12$$22$$264$$P$$7$$7p$$20$$4$$80$$25$$2$$50$$N=50$$\sum{fx=524+7p}$ We know that, Mean...
for the given distribution whose 
is mean$x:$$8$$12$$15$$p$$20$$25$$30$$f:$$12$$16$$20$24$16$$8$$4$ Solution: $x$$f$$fx$$8$$12$$96$$12$$16$$192$$15$$20$$300$$P$$24$$24p$$20$$16$$320$$25$$8$$200$$30$$4$$120$$N=100$$\sum{fx=1228+24p}$ We...
Find the median height. Solution: Here, we have N = \[420,\] So\[,\text{ }N/2\text{ }=\text{ }420/\text{ }2\text{ }=\text{ }210\] The cumulative frequency just greater than \[N/2\text{ }is\text{...
, find 
$x:$$5$10$15$$20$$25$$f:$$6$$p$$6$$10$$5$ Solution: $x$$f$$fx$$5$$6$$30$$10$$p$$10p$$15$$6$$90$$20$$10$$200$$25$$5$$125$$N=p+27$$\sum{fx=445+10p}$ We know that, Mean $=\sum{fx/N=\left( 445+10p...
is the mean of given data. Find the value of $x:$$10$$15$$p$$25$$35$$f:$$3$$10$$25$$7$$5$ Solution: $x$$f$$fx$$10$$3$$30$$15$$10$$150$$p$$25$$25p$$25$$7$$175$$35$$5$$175$$N=50$$\sum{fx=530+25p}$ We know that, Mean $=\sum{fx/N=\left( 2620+25p...
$x:$$19$$21$$23$$25$$27$$29$$31$$f:$$13$$15$$16$$18$$16$$15$$13$ Solution: $x$$f$$fx$$19$$13$$247$$21$$15$$315$$23$$16$$368$$25$$18$$450$$27$$16$$432$$29$$15$$435$$31$$13$$403$$N=106$$\sum{fx}=2620$...
Age (in years)Number of policy holderBelow 202Below 256Below 3024Below 3545Below 4078Below 4589Below 5092Below 5598Below 60100 Solution: Class interval ...
Class IntervalFrequency0-10510-20x20-302030-401540-50y50-605Total60 Arrangement: Given information, n = 60 Middle of the given information = 28.5 Where, n/2 = 30 Middle class is 20 – 30 with an...
Monthly consumption(in units)No. of customers65-85485-1055105-12513125-14520145-16514165-1858185-2054 Solution: Track down the aggregate recurrence of the given information as follows: Class...
Number of carsFrequency0-10710-201420-301330-401240-502050-601160-701570-808 solution Given Data: Modular class = 40 – 50, l = 40, Class width (h) = 10, fm = 20, f1 = 12 and f2 = 11 \[Mode\text{...
Run ScoredNumber of Batsman3000-400044000-5000185000-600096000-700077000-800068000-900039000-10000110000-110001 Find the mode of the data. Solution: Given information: Modular class = 4000 – 5000, l...
No of Students per teacherNumber of states / U.T15-20320-25825-30930-351035-40340-45045-50050-552 Solution: Given information: Modular class = 30 – 35, l = 30, Class width (h) = 5, fm = 10, f1 = 9...
ExpenditureNumber of families1000-1500241500-2000402000-2500332500-3000283000-3500303500-4000224000-4500164500-50007 Solution: From the given information the modular class is 60–80. l = 60, The...
Lifetime (in hours)0-2020-4040-6060-8080-100100-120Frequency103552613829 Determine the modal lifetimes of the components. Solution: From the given information the modular class is 60–80. l = 60, The...
Age (in years)5-1515-2525-3535-4545-5555-65Number of patients6112123145 Find the mode and the mean of the data given above. Compare and interpret the twomeasures of central tendency. Solution: To...
Literacy rate (in %)45-5555-6565-7575-8585-98Number of cities3101183 Solution: Discover the midpoint of the given stretch utilizing the recipe. Midpoint (xi) = (maximum cutoff + lower limit)/2 For...
Arrangement: Discover the midpoint of the given stretch utilizing the equation. Midpoint (xi) = (maximum cutoff + lower limit)/2 Class interval Frequency (fi) Mid-point (xi) fixi 0-6 11 3 33 6-10 10...
Concentration of SO2 ( in ppm)Frequency0.00 – 0.0440.04 – 0.0890.08 – 0.1290.12 – 0.1620.16 – 0.2040.20 – 0.242 Find the mean concentration of SO2 in the air. SOLUTION To discover the mean,...
Daily expenditure(in c)100-150150-200200-250250-300300-350Number of households451222 Solution: Discover the midpoint of the given span utilizing the recipe. \[Midpoint\text{ }\left( xi \right)\text{...
Number of mangoes50-5253-5556-5859-6162-64Number of boxes1511013511525 Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose? Solution: Since, the...
solution: From the given information, let us expect the mean as A = 75.5 \[xi\text{ }=\text{ }\left( Upper\text{ }cutoff\text{ }+\text{ }Lower\text{ }limit \right)/2\] Class size (h) = 3 Presently,...
.solution: To discover the missing recurrence, utilize the mean equation. Here, the worth of mid-point $\left(\mathrm{x}_{\mathrm{i}}\right)$ mean $\overline{\mathrm{x}}=18$
and 
and 
and 
and 
\14 8 ( 10 Solution: Discover the midpoint of the given span utilizing the recipe. Midpoint $\left(\mathrm{x}_{\mathrm{i}}\right)=($ maximum breaking point $+$ lower limit $)/2$ For this situation,...
Solution: From the offered information, to address the table as chart, pick the maximum furthest reaches of the class stretches are in $x$-hub and frequencies on $\mathrm{y}$-pivot by picking the...
Solution: Convert the given appropriation table to a not as much as type aggregate recurrence circulation, and we get Every day income Frequency Cumulative Frequency Under 120 12 12 Under 140 14 26...
Solution: Changing the given dissemination over to a more than type conveyance, we get Creation Yield (kg/ha) Number of ranches More than or equivalent to 50 100 More than or equivalent to 55 100-2...
Solution: To track down the mean worth, we will utilize direct technique in light of the fact that the mathematical worth of fi and xi are little. Discover the midpoint of the given stretch...