The mean of five observations is $15$. If the mean of first three observations is $14$ and that of the last three observations is $17$, then the third observation is
Assertion (A) : Median marks of students in a class test is $16$. It means half of the class got marks less than $16$. Reason (R) : Median divides the distribution in two equal parts.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
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(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Assertion (A): Median of a data is the value of $\frac{N}{2}$, where $N$ represents sum of all frequencies. Reason (R): Median divides the whole distribution in two equal parts.
(a)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
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Answer (D) Assertion (A) is false, but Reason (R) is true.
After an examination, a teacher wants to know the marks obtained by maximum number of the students in her class. She requires to calculate ______ of marks.
Let 'a' be the assumed mean and 'h' be the class size for a grouped data. Which of the following is not the correct formula to find the mean of the grouped data ?
(a)$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$
(b)$\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$, where $d_i = x_i - a$
(c)$\bar{x} = a + \frac{\Sigma f_i u_i}{\Sigma f_i}$, where $u_i = \frac{x_i - a}{h}$
(d)$\bar{x} = a + \frac{\Sigma f_i y_i}{\Sigma f_i} \times h$, where $y_i = \frac{x_i - a}{h}$
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(C) $\bar{x} = a + \frac{\Sigma f_i u_i}{\Sigma f_i}$, where $u_i = \frac{x_i - a}{h}$
Find the mean and the median for the following frequency distribution : Class $11-13$ $13-15$ $15-17$ $17-19$ $19-21$ $21-23$ $23-25$ Frequency $7$ $6$ $9$ $13$ $20$ $5$ $4$
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Calculation of Mean: $\sum f_i = 64$ $\sum f_i x_i = 1152$ Mean $= \frac{\sum f_i x_i}{\sum f_i} = \frac{1152}{64} = 18$. Calculation of Median: $N = 64 \Rightarrow \frac{N}{2} = 32$. The median class is $17-19$ (since cumulative frequency $22 < 32 < 35$). Lower limit of median class $l = 17$. Cumulative frequency of class preceding median class $cf = 22$. Frequency of median class $f = 13$. Class size $h = 2$. Median $= l + (\frac{\frac{N}{2} - cf}{f}) \times h$ Median $= 17 + (\frac{32 - 22}{13}) \times 2$ Median $= 17 + (\frac{10}{13}) \times 2 = 17 + \frac{20}{13}$ Median $= 17 + 1.538 \approx 18.54$.
Activities like running or cycling reduce stress and the risk of mental disorders like depression. Running helps build endurance. Children develop stronger bones and muscles and are less prone to gain weight. The physical education teacher of a school has decided to conduct an inter school running tournament in his school premises. The time taken by a group of students to run $100 \text{ m}$, was noted as follows : Based on the above, answer the following questions : (i) What is the median class of the above given data ? (ii) (a) Find the mean time taken by the students to finish the race. OR (b) Find the mode of the above given data. (iii) How many students took time less than $60$ seconds ?
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Time (in seconds) | Number of students $0-20$ | $8$ $20-40$ | $10$ $40-60$ | $13$ $60-80$ | $6$ $80-100$ | $3$ Total | $40$ (i) Correct Cummulative Frequency Median class = $40-60$ (ii) (a) Correct table for $x_i$ and $f_ix_i$ Time (in sec) | Number of students (f) | $x_i$ | cf | $f_ix_i$ $0-20$ | $8$ | $10$ | $8$ | $80$ $20-40$ | $10$ | $30$ | $18$ | $300$ $40-60$ | $13$ | $50$ | $31$ | $650$ $60-80$ | $6$ | $70$ | $37$ | $420$ $80-100$ | $3$ | $90$ | $40$ | $270$ Total | $40$ | | | $1720$ Mean = $$\begin{aligned}& \frac{1720}{40} = 43\ \text{OR} \\ & (b) \text{ Modal class } = 40-60 \\ & \text{Mode } = 40 + \frac{(13-10)}{(26-10-6)} \times 20 \\ & = 46 \\ & (iii)\end{aligned}$$31 students took time less than 60 seconds
CENTRAL POLLUTION CONTROL BOARD'S AIR QUALITY STANDARDS AIR QUALITY INDEX (AQI) CATEGORY 0-50 Good 51-100 Satisfactory 101-200 Moderate 201-300 Poor 301-400 Very Poor 401-500 Severe The Air Quality Index (AQI) is a scale from $0$ to $500$ that indicates air quality, with higher numbers signifying more pollution and greater health concerns. Mansi collected the daily data of AQI of her city for a month and presented it as given below: AQI Range : 1-100 101-200 201-300 301-400 401-500 Number of Days: $3$ $9$ $12$ $4$ $2$ (i) Convert the data to continuous frequency distribution. (ii) What is the quality of air in most of the days of the month ? (iii) (a) Using table formed in part (i), find mode of the data. OR (b) Using table formed in part (i), find median of the data.
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(i) Class Interval frequency 0.5-100.5 3 100.5 - 200.5 9 200.5 - 300.5 12 300.5-400.5 4 400.5-500.5 2 (1) (ii) Air Quality is poor on most of the days (1) (iii) (a) Modal class is $200.5 - 300.5$ (½) Mode = $200.5 + 100 (\frac{12-9}{2\times 12-9-4})$ (1) $= 227.77$ (½) OR (b) Class Interval frequency cf 0.5-100.5 3 3 100.5 - 200.5 9 12 200.5 - 300.5 12 24 300.5-400.5 4 28 400.5-500.5 2 30 For correct table (½) Median class is $200.5 - 300.5$ (½) Median = $200.5 + (\frac{15-12}{12}) \times 100$ (1) $= 225.5$ (½)
CENTRAL POLLUTION CONTROL BOARD'S AIR QUALITY STANDARDS AIR QUALITY INDEX (AQI) CATEGORY 0-50 Good 51-100 Satisfactory 101-200 Moderate 201-300 Poor 301-400 Very Poor 401-500 Severe The Air Quality Index (AQI) is a scale from $0$ to $500$ that indicates air quality, with higher numbers signifying more pollution and greater health concerns. Mansi collected the daily data of AQI of her city for a month and presented it as given below : AQI Range: $1-100$ $101-200$ $201-300$ $301-400$ $401-500$ Number of Days: $3$ $9$ $12$ $4$ $2$ (i) Convert the data to continuous frequency distribution. (ii) What is the quality of air in most of the days of the month? (iii) (a) Using table formed in part (i), find mode of the data. OR (b) Using table formed in part (i), find median of the data.
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(i) Class Interval frequency $0.5-100.5$ $3$ $100.5-200.5$ $9$ $200.5-300.5$ $12$ $300.5-400.5$ $4$ $400.5-500.5$ $2$ (1 Mark) (ii) Air Quality is poor on most of the days (1 Mark) (iii) (a) Modal class is $200.5 - 300.5$ (1/2 Mark) Mode = $200.5 + 100 (\frac{12-9}{2\times12-9-4})$ (1 Mark) = $227.77$ (1/2 Mark) OR (b) Class Interval frequency cf $0.5-100.5$ $3$ $3$ $100.5-200.5$ $9$ $12$ $200.5-300.5$ $12$ $24$ $300.5-400.5$ $4$ $28$ $400.5-500.5$ $2$ $30$ For correct table (1/2 Mark) Median class is $200.5 - 300.5$ (1/2 Mark) Median = $200.5 + (\frac{15-12}{12}) \times 100$ (1 Mark) = $225.5$ (1/2 Mark)
A survey regarding the heights (in cm) of $50$ girls of class X of a school was conducted and the following data was obtained : Find the mean and mode of the above data.
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Correct table Mean $= 145 + \frac{24}{50} \times 10$ $= 149.8$ $\therefore$ mean height is $149.8$ cm Modal class is $150 - 160$ Mode $= 150 + \frac{(20-12)}{(2\times20-12-8)} \times 10$ $= 154$ $\therefore$ modal height is $154$ cm
A student noted the number of cars passing through a spot on a road for $100$ periods each of $3$ minutes and summarised it in the table given below. Find the mean and median of the following data. Number of cars Frequency (periods) 0-10 7 10-20 14 20-30 13 30-40 12 40-50 20 50-60 11 60-70 15 70-80 8
A student noted the number of cars passing through a spot on a road for $100$ periods each of $3 \text{ minutes}$ and summarised it in the table given below. Find the mean and median of the following data.
Find the Mean and Mode of the following data : Class $4-8$ $8-12$ $12-16$ $16-20$ $20-24$ $24-28$ $28-32$ $32-36$ Frequency $2$ $12$ $15$ $25$ $18$ $12$ $13$ $3$
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Correct table ($1\frac{1}{2}$) Mean $= 22 + \frac{(-52)}{100} \times 4$ ($1$) $= 19.92$ ($1/2$) Modal Class is $16 - 20$ ($1/2$) Mode $= 16 + \left(\frac{25-15}{2\times25-15-18}\right) \times 4$ ($1\frac{1}{2}$) $= \frac{312}{17}$ or $18.35$ approx.
During a medical checkup, height of 35 students of a class were recorded as follows :
Height (in cm)
90-100
100-110
110-120
120-130
130-140
140-150
Number of Students
3
2
4
5
14
7
Find the difference between the mean height and median height.
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Correct table Mean $= 115 + \frac{46}{35} \times 10 = \frac{897}{7}$ or 128.14 approx. $\therefore$ Mean height is $\frac{897}{7}$ cm or 128.14 cm approx. Median Class is 130 - 140 Median $= 130 + \frac{\frac{35}{2} - 14}{14} \times 10 = 132.5$ $\therefore$ Median height is 132.5 cm Difference of mean height and median height $= 132.5 - 128.14 = 4.36$ cm
An SBI health insurance agent found the following data for distribution of ages of $100$ policy holders. The health insurance policies are given to persons of age $15$ years and onwards, but less than $60$ years. Age (in yrs) $15-20$ $20-25$ $25-30$ $30-35$ $35-40$ $40-45$ $45-50$ $50-55$ $55-60$ Number of policy holders $2$ $4$ $18$ $21$ $33$ $11$ $3$ $6$ $2$ Find the modal age and median age of the policy holders.
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For correct table Age (in years) f cf $15-20$ $2$ $2$ $20-25$ $4$ $6$ $25-30$ $18$ $24$ $30-35$ $21$ $45$ $35-40$ $33$ $78$ $40-45$ $11$ $89$ $45-50$ $3$ $92$ $50-55$ $6$ $98$ $55-60$ $2$ $100$ Total $100$ (1 Mark) Modal class = $35-40$ (1/2 Mark) Mode $= 35 + \frac{33-21}{2(33)-21-11} \times 5$ (1 Mark) $= \frac{625}{17} = 36.7$(approx.) (1/2 Mark) $\therefore$ Modal age = $36.7$ years (approx.) $\frac{n}{2} = 50$, Median class = $35- 40$ (1/2 Mark) Median $= 35 + \frac{50-45}{33} \times 5$ (1 Mark) $= \frac{1180}{33} = 35.7$(approx.) (1/2 Mark) $\therefore$ Median age = $35.7$ years (approx.)
An SBI health insurance agent found the following data for distribution of ages of 100 policy holders. The health insurance policies are given to persons of age 15 years and onwards, but less than 60 years.
Age (in yrs)
Number of policy holders
15 - 20
2
20 - 25
4
25 - 30
18
30 - 35
21
35 - 40
33
40 - 45
11
45 - 50
3
50 - 55
6
55 - 60
2
Find the modal age and median age of the policy holders.
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Age (in years) | f | cf 15 - 20 | 2 | 2 20 - 25 | 4 | 6 25 - 30 | 18 | 24 30 - 35 | 21 | 45 35 - 40 | 33 | 78 40 - 45 | 11 | 89 45 - 50 | 3 | 92 50 - 55 | 6 | 98 55 - 60 | 2 | 100 Total | 100 For correct table (1 Mark) Modal class = 35-40 (1/2 Mark) Mode = $35 + \frac{33-21}{2(33)-21-11} \times 5$ (1 Mark) $= \frac{625}{17} = 36.7(\text{approx.})$ (1/2 Mark) Modal age = 36.7 years (approx.) $\frac{n}{2} = 50$, Median class = 35 – 40 (1/2 Mark) Median = $35 + \frac{50-45}{33} \times 5$ (1 Mark) $= \frac{1180}{33} = 35.7(\text{approx.})$ (1/2 Mark) Median age = 35.7 years (approx.)
A life insurance agent found the following data for the distribution of $100$ policy holders on the basis of their ages. begin{tabular}{|c|c|} hline Age (in years) & Number of policy holders hline $15-20$ & $2$ hline $20-25$ & $4$ hline $25-30$ & $18$ hline $30-35$ & $21$ hline $35-40$ & $33$ hline $40-45$ & $11$ hline $45-50$ & $3$ hline $50-55$ & $6$ hline $55-60$ & $2$ hline end{tabular} Find the median age of the policy holders.
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begin{tabular}{|c|c|c|} hline $CI$ & $f_i$ & $Cf$ hline $15-20$ & $2$ & $2$ hline $20-25$ & $4$ & $6$ hline $25-30$ & $18$ & $24$ hline $30-35$ & $21$ & $45$ hline $35-40$ & $33$ & $78$ hline $40-45$ & $11$ & $89$ hline $45-50$ & $3$ & $92$ hline $50-55$ & $6$ & $98$ hline $55-60$ & $2$ & $100$ hline end{tabular} $\frac{N}{2} = 50 \therefore$ median class : $35-40$ Median $= 35 + \frac{50 - 45}{33} \times 5 = 35.76$ Thus, the median age of the policy holders is $35.76$ years.
A class teacher has the following absentees record of $30$ students of a class. begin{tabular}{|c|c|c|c|c|c|c|} hline Number of days & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 & 20 - 24 hline Number of Absent students & 1 & 8 & x & 6 & 5 & y hline end{tabular} If the mean number of days a student was absent is $12$, find the values of $x$ and $y$.
Find 'mean' and 'mode' of the following data : begin{tabular}{|l|c|c|c|c|c|c|} hline Class & 10-25 & 25-40 & 40-55 & 55-70 & 70-85 & 85-100 hline Number of Students & 12 & 10 & 15 & 13 & 8 & 12 hline end{tabular}
Find the mode and the mean of the following frequency distribution : begin{tabular}{|l|c|c|c|c|c|} hline Class & $0-8$ & $8-16$ & $16-24$ & $24-32$ & $32-40$ hline Frequency & $6$ & $7$ & $10$ & $8$ & $9$ hline end{tabular}
Weekly expenditure on Ayurvedic medicines of few households in a locality is recorded below. If the mean expenditure for this is ₹211, then find the value of the missing frequency 'y'.
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Correct table Mean $= 211$ $225 + \frac{(-7)}{13+y} \times 50 =211$ $\Rightarrow y = 12$
The monthly expenditure on milk in $200$ families of a Housing Society is given below : Monthly Expenditure (in ₹) $1000-1500$ $1500-2000$ $2000-2500$ $2500-3000$ $3000-3500$ $3500-4000$ $4000-4500$ $4500-5000$ Number of families $24$ $40$ $33$ $x$ $30$ $22$ $16$ $7$ Find the value of $x$ and also, find the median and mean expenditure on milk.
250 apples of a box were weighed and the distribution of masses of the apples is given in the following table : Mass (in grams) 80-100 100-120 120-140 140-160 160-180 Number of apples 20 60 70 x 60 (i) Find the value of $x$ and the mean mass of the apples. (ii) Find the modal mass of the apples.
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(i) $20 + 60 +70 + x + 60 = 250$ $x = 250-210 = 40$ Mass 80-100 100-120 120-140 140-160 160-180 Total No. of apples $f_i$ 20 60 70 $x = 40$ 60 250 $x_i$ 90 110 130 150 170 $f_ix_i$ 1800 6600 9100 6000 10200 33700 Mean mass $= \frac{33700}{250} = 134.8$ Mean mass = $134.8$ g (ii) Modal class = 120-140 Mode $= 120+\frac{(70-60)}{(140-60-40)} \times 20$ $= 125$ Hence modal mass = $125$ gm
250 apples of a box were weighed and the distribution of masses of the apples is given in the following table : Mass (in grams) Number of apples (i) Find the value of $x$ and the mean mass of the apples. (ii) Find the modal mass of the apples
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(i)$20 + 60 + 70 + x + 60 = 250$ $x = 250-210 = 40$ Mass No. of apples $f_i$ $x_i$ $x_if_i$ Mean mass = $\frac{33700}{250} = 134.8$ Mean mass = $134.8$ g (ii) Modal class = $120-140$ Mode = $120 + \frac{(70-60)}{(140-60-40)} \times 20$ = $125$ Hence modal mass = $125$ g
An age-wise list of number of literate people in a block is prepared in the following table. There are total $100$ people and their median age is $41.5$ years. Information about two groups are missing, which are denoted by $x$ and $y$. Find the value of $x$ and $y$. Age (in years) Number of literate people $10-20$ $15$ $20-30$ $x$ $30-40$ $12$ $40-50$ $20$ $50-60$ $y$ $60-70$ $8$ $70-80$ $10$
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Age (in years) Number of literate people ($f_i$) Cumulative frequency $10-20$ $15$ $15$ $20-30$ $x$ $15 + x$ $30-40$ $12$ $27 + x$ $40-50$ $20$ $47 + x$ $50-60$ $y$ $47 + x + y$ $60-70$ $8$ $55 + x + y$ $70-80$ $10$ $65 + x + y$ $65 + x + y = 100$ $\Rightarrow x + y = 35$ ...(i) Median $= 41.5$ $40-50$ is the median class. $\Rightarrow 41.5 = 40 + \frac{\frac{100}{2}-27-x}{20} \times 10$ Solving, we get $x = 20$ From (i), $y = 15$
Mode of the following $30$ observations is $175$. Find the values of the missing frequencies x and y. Class Interval Frequency $0-50$ $4$ $50-100$ $3$ $100-150$ $5$ $150-200$ $x$ $200-250$ $y$ $250-300$ $3$ $300-350$ $4$
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Here, modal class $= 150 - 200$ and $f_0 = 5, f_1 = x, f_2 = y$ and $h = 50$ Mode $= 175$ $150 + \frac{x-5}{2x-5-y} \times 50 = 175$ $\Rightarrow y = 5$ Also, $19 + x + y = 30$ $\Rightarrow x = 6$
Assertion (A): If the Mode and Mean of a data are $12 k$ and $15 k$, then Median of the data is $14 k$. Reason (R) : The relation between the Mean, Mode and Median of a data is: Mean = $3$ Median $- 2$ Mode.
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(C) Assertion (A) is true, but Reason (R) is false.
Assertion (A) : If the difference of the mode and median of a data is $24$, then the difference of the median and mean is $12$. Reason (R) : $\text{Mode} = 3 \text{ mean} - 2 \text{ median}$.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
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(C) Assertion (A) is true, but Reason (R) is false.
Assertion (A): If the value of mode and mean for a distribution is $50$ and $56$ respectively, then the value of median is $54$. Reason (R) : Median = $\frac{1}{3}$ (Mode - $2$ Mean)
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(C) Assertion (A) is true, but Reason (R) is false.
BINGO is game of chance. The host has $75$ balls numbered $1$ through $75$. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game. The table given below, shows the data of one such game where $48$ balls were used before Tara said 'BINGO'.
Numbers announced
Number of times
0-15
8
15-30
9
30-45
10
45-60
12
60-75
9
Based on the above information, answer the following : (i) Write the median class. (ii) When first ball was picked up, what was the probability of calling out an even number? (iii) (a) Find median of the given data. OR (b) Find mode of the given data.
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Number announced | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 Number of times (f) | 8 | 9 | 10 | 12 | 9 cf | 8 | 17 | 27 | 39 | 48=N (i) $\frac{N}{2} = 24$ $\therefore$ median class is $30 - 45$ (ii) P (picking up an even number) = $\frac{37}{75}$ (iii) (a) Median = $30 + \frac{\left(\frac{48}{2} - 17\right)}{10} \times 15$ $= 40.5$ OR (iii) (b) Modal class is $45 - 60$ Mode = $45 + \frac{12-10}{2\times 12-10-9} \times 15$ $= 51$
The denominator of a fraction is $2$ more than the numerator. If $2$ is added to both its numerator and denominator, then the sum of the new fraction and the original fraction is $\frac{46}{35}$. Find the original fraction.
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Let the fraction be $\frac{x}{x+2}$ Therefore, $\frac{x}{x+2} + \frac{x+2}{x+4} = \frac{46}{35}$ $\Rightarrow 24x^2 + 4x - 228 = 0$ or $6x^2 + x - 57 = 0$ $\Rightarrow (6x + 19)(x - 3) = 0$ $x \neq -\frac{19}{6}$ $\therefore x = 3$ So, the required fraction is $\frac{3}{5}$
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