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
The mean of seven observations is $17$. If the mean of the first four observations is $15$ and that of the last four observations is $18$, then the fourth observation is :
Following data shows the marks obtained by $100$ students in a class test : Marks obtained: $20, 29, 28, 33, 42, 38, 43, 25$. Number of students: $6, 28, 24, 15, 2, 4, 1, 20$. The median will be the average of which two observations ?
The median of a set of $15$ distinct observations is $30.5$. If each of the largest $7$ observations of the set is increased by $3$, then the median of the new set.
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.
The distribution below gives the marks obtained by 80 students on a test : Marks Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Number of Students 3 12 27 57 75 80 The modal class of this distribution is :
In the following frequency distribution : Height (in cm) : $120-125$ $125-130$ $130-135$ $135-140$ $140-145$ Number of students : $17$ $12$ $13$ $8$ $10$ the sum of the upper limit of the modal class and the lower limit of the median class is :
For some data $x_1, x_2, \dots, x_n$ with respective frequencies $f_1, f_2, \dots, f_n$, the value of $\sum_{i=1}^{n} f_i (x_i - \bar{x})$ is equal to :
For the above distribution, the modal class is : Marks : Below $10$ Below $20$ Below $30$ Below $40$ Below $50$ Number of Students : $3$ $12$ $27$ $57$ $75$
Following data shows the marks obtained by 100 students in a class test : Marks obtained | 20 | 29 | 28 | 33 | 42 | 38 | 43 | 25 Number of students | 6 | 28 | 24 | 15 | 2 | 4 | 1 | 20 The median will be the average of which two observations ?
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 $7$ $6$ $9$ $13$ $20$ $5$ $4$
Show SolutionHide Solution↓
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 $= 17 + 1.538 \approx 18.54$.
Find the mean of the following frequency distribution : Classes quad $25-30$ quad $30-35$ quad $35-40$ quad $40-45$ quad $45-50$ quad $50-55$ quad $55-60$ Frequency quad $14$ quad $22$ quad $16$ quad $6$ quad $5$ quad $3$ quad $4$
Find the mean of the following frequency distribution : Classes $25-30$ $30-35$ $35-40$ $40-45$ $45-50$ $50-55$ $55-60$ Frequency $14$ $22$ $16$ $6$ $5$ $3$ $4$
Show SolutionHide Solution↓
For correct table $2$ Marks Mean $= 42.5 - \frac{79}{70} \times 5 = 36.86$
The government rescued $100$ people after a train accident. Their ages were recorded in the following table. Find their mean age. Age (in years) Number of people rescued $10-20$ $9$ $20-30$ $14$ $30-40$ $15$ $40-50$ $21$ $50-60$ $23$ $60-70$ $12$ $70-80$ $6$
Show SolutionHide Solution↓
Age (in years) Number of people rescued ($f_i$) $x_i$ $u_i$ $f_iu_i$ $10-20$ $9$ $15$ $-3$ $-27$ $20-30$ $14$ $25$ $-2$ $-28$ $30-40$ $15$ $35$ $-1$ $-15$ $40-50$ $21$ $45$ $0$ $0$ $50-60$ $23$ $55$ $1$ $23$ $60-70$ $12$ $65$ $2$ $24$ $70-80$ $6$ $75$ $3$ $18$ Total $100$ $-5$ Mean Age $= 45 + \frac{(-5)}{100} \times 10$ $= 44.5$ Hence, mean age is $44.5$ years
The table below shows the daily expenditure on food of $25$ households in a locality : Daily expenditure (in ₹) : $200-250$ $250-300$ $300-350$ $350-400$ $400-450$ Number of households : $4$ $5$ $12$ $2$ $2$ Find the mean daily expenditure on food.
Show SolutionHide Solution↓
Sol. Correct table Mean = $325 + \frac{(-7)}{25} \times 50$ = $311$ Therefore, the mean daily expenditure on food is ₹311.
In a test, the marks obtained by $100$ students (out of $50$) are given below : Marks obtained : $0-10 \quad 10-20 \quad 20-30 \quad 30-40 \quad 40-50$ Number of students : $12 \quad 23 \quad 34 \quad 25 \quad 6$ Find the mean marks of the students.
Show SolutionHide Solution↓
Marks for correct table: $1\frac{1}{2}$ Marks Obtained: $0-10$, $f_i=12$, $x_i=5$, $f_ix_i=60$ Marks Obtained: $10-20$, $f_i=23$, $x_i=15$, $f_ix_i=345$ Marks Obtained: $20-30$, $f_i=34$, $x_i=25$, $f_ix_i=850$ Marks Obtained: $30-40$, $f_i=25$, $x_i=35$, $f_ix_i=875$ Marks Obtained: $40-50$, $f_i=6$, $x_i=45$, $f_ix_i=270$ Total: $f_i=100$, $f_ix_i=2400$ Mean $= \frac{2400}{100}$ $= 24$
India meteorological department observes seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to compare and analyse the results. The table given below shows sub-division wise seasonal (monsoon) rainfall (mm) in 2018: Rainfall (mm) Number of Sub-divisions 200-400 2 400-600 4 600-800 7 800-1000 4 1000-1200 2 1200-1400 3 1400-1600 1 1600-1800 1 Based on the above information, answer the following questions: (I) Write the modal class. (II) Find the median of the given data.
Show SolutionHide Solution↓
(i) Modal Class is 600-800 (ii) $\frac{N}{2} = 12$, median class is 600 – 800 Rainfall $x_i$ $f_i$ cf. 200-400 300 2 2 400-600 500 4 6 600-800 700 7 13 800-1000 900 4 17 1000 - 1200 1100 2 19 1200-1400 1300 3 22 1400-1600 1500 1 23 1600-1800 1700 1 24 24 Median = $600 + \frac{200}{7} (12-6)$ or $771-4$ OR (ii) Rainfall $X_i$ $f_i$ $f_i x_i$ 200-400 300 2 600 400-600 500 4 2000 600-800 700 7 4900 800-1000 900 4 3600 1000 - 1200 1100 2 2200 1200-1400 1300 3 3900 1400-1600 1500 1 1500 1600-1800 1700 1 1700 24 20400 Mean = $\frac{20400}{24} = 850$ (iii) Sub-divisions having good rainfall = $2 + 3 + 1 + 1 = 7$.
Case Study – 3 A survey was conducted by the Education Ministry of India to record the teacher-student ratio in various higher secondary schools of India. The following distribution was given by the Ministry : Number of students/teacher : $15-20$ $20-25$ $25-30$ $30-35$ $35-40$ $40-45$ Number of Schools : $3$ $8$ $9$ $10$ $3$ $2$ Based on the above information, answer the following questions : (i) Write the modal class. (ii) Write the median class. (iii) (a) Find the mode of the data. OR (b) Find the median of the data.
Show SolutionHide Solution↓
Sol. (i) Modal class is $30 - 35$. (ii) Median class is $25 - 30$. (iii) (a) Mode = $30 + \frac{(10-9)}{(2 \times 10-9-3)} \times 5$ = $30.625$ OR (b) Median = $25 + \frac{(\frac{35}{2}-11)}{9} \times 5$ = $28.61$ approx.
Vocational training complements traditional education by providing practical skills and hands-on experience. While education equips individuals with a broad knowledge base, vocational training focuses on job-specific skills, enhancing employability thus making the student self-reliant. Keeping this in view, a teacher made the following table giving the frequency distribution of students/adults undergoing vocational training from the training institute. From the above answer the following questions: (i) What is the lower limit of the modal class of the above data? (ii) (a) Find the median class of the above data. OR (b) Find the number of participants of age less than $50$ years who undergo vocational training. (iii) Give the empirical relationship between mean, median and mode.
Show SolutionHide Solution↓
(i) Modal class is $19.5 - 24.5$ Lowe limit $=19.5$ (ii) (a) Correct table $\frac{n}{2} = \frac{365}{2} = 182.5$ median class = $19.5 - 24.5$ OR (ii) (b) $62 + 132 + 96 + 37 + 13 + 11 + 10 = 361$ (iii) $3\text{median} = \text{mode} + 2 \text{ mean}$
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 ?
Show SolutionHide Solution↓
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$
Case Study – 1 Student-teacher ratio expresses the relationship between the number of students enrolled in a school and the number of teachers employed by the school. This ratio is important for a number of reasons. It can be used as a tool to measure teachers' workload as well as the allocation of resources. A survey was conducted in $100$ secondary schools of a state and the following frequency distribution table was prepared : Number of students per Teacher: $20-25$, $25-30$, $30-35$, $35-40$, $40-45$, $45-50$ Number of Schools: $5$, $15$, $25$, $30$, $15$, $10$ Based on the above, answer the following questions : (i) What is the lower limit of the median class? (ii) What is the upper limit of the modal class ? (iii) (a) Find the median of the data. OR (b) Find the modal of the data.
Show SolutionHide Solution↓
No. of Students per teacher | No. of School | c.f. $20-25$ | $5$ | $5$ $25-30$ | $15$ | $20$ $30-35$ | $25$ | $45$ $35-40$ | $30$ | $75$ $40-45$ | $15$ | $90$ $45-50$ | $10$ | $100$ (i) Median class is $35 - 40$ Lower limit of median class = $35$ (ii) Modal class is $35 - 40$ Upper limit of modal class = $40$ (iii) (a) Median class is $35 - 40$ Median = $$\begin{aligned}& 35 + \frac{(\frac{100}{2} - 45)}{30} \times 5 \\ & = \frac{215}{6}\end{aligned}$$ or $35.83$ approx. OR (iii) (b) Modal class is $35 - 40$ Mode = $$\begin{aligned}& 35 + \frac{30-25}{2\times30-25-15} \times 5 \\ & = 36.25\end{aligned}$$
The India Meteorological Department observes seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to compare and analyse the results. The table below shows sub-divisions wise seasonal (monsoon) rainfall (in mm) in 2023. Rainfall (mm) | No. of Sub-divisions 200-400 | 3 400-600 | 4 600-800 | 7 800-1000 | 4 1000-1200 | 3 1200-1400 | 3 Based on the information given above, answer the following questions: (i) Write the modal class. (ii) (a) Find the median of the given data. OR (b) Find the mean rainfall in the season. (iii) If a sub-division having at least $800 \operatorname{mm}$ rainfall during monsoon season is considered a good rainfall sub-division, then how many sub-divisions had good rainfall?
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.
Show SolutionHide Solution↓
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 survey regarding the heights (in cm) of $50$ girls of class X of a school was conducted and the following data was obtained: Height (in cm) | Number of girls 120-130 | 2 130-140 | 8 140-150 | 12 150-160 | 20 160-170 | 8 Total | 50 Find the mean and mode of the above data.
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.
In a test, the marks obtained by 100 students (out of 50) are given below : Marks obtained: 0-10 10-20 20-30 30-40 40-50 Number of students : 12 23 34 25 6 Find the mean marks of the students.
The following table shows the ages of the patients admitted in a hospital during a year : Age (in years) $5-15$ $15-25$ $25-35$ $35-45$ $45-55$ $55-65$ Number of patients $6$ $11$ $21$ $23$ $14$ $5$ Find the mode and mean of the data given above.
Show SolutionHide Solution↓
Correct table Age (in years) No. of patients ($f_i$) Mid point ($x_i$) $x_if_i$ $5-15$ $6$ $10$ $60$ $15-25$ $11$ $20$ $220$ $25-35$ $21$ $30$ $630$ $35-45$ $23$ $40$ $920$ $45-55$ $14$ $50$ $700$ $55-65$ $5$ $60$ $300$ Total $80$ $$\begin{aligned}& 2830 \\ & \Rightarrow \text{Mean } = \frac{2830}{80} \\ & = 35.375 \\ & \text{Modal class } = (35 - 45) \\ & \Rightarrow \text{Mode } = 35 + (\frac{23-21}{2\times23-21-14}) \times h \\ & = 36.81 \\ & \text{Therefore, mode and mean of given data are } 36.81 \text{ years and } 35.375 \text{ years respectively.}\end{aligned}$$
The following table shows the ages of the patients admitted in a hospital during a year : Find the mode and mean of the data given above.
Show SolutionHide Solution↓
Correct table $\Rightarrow \text{Mean} = \frac{2830}{80}$ $= 35.375$ Modal class = $(35 - 45)$ $\Rightarrow \text{Mode} = 35 + \left(\frac{23-21}{2\times23-21-14}\right) \times h$ $= 36.81$ Therefore, mode and mean of given data are $36.81$ years and $35.375$ years respectively.
The following frequency distribution gives the monthly consumption of electricity of $68$ consumers of a locality. Find the mean and mode of the data: Monthly Consumption (in units) Number of Consumers $65-85$ $4$ $85-105$ $5$ $105-125$ $13$ $125-145$ $20$ $145-165$ $14$ $165-185$ $8$ $185-205$ $4$
Show SolutionHide Solution↓
Correct Table (1 frac{1}{2}) Mean = $135+\frac{7}{68} \times 20 = 137.06$ (1) Modal Class is $125-145$ (1/2) Mode = $125 + \left(\frac{20-13}{40-13-14}\right) \times 20 = 135.77$ (1/2) Hence, Mean = $137.06$ units and Mode = $135.77$ units (1/2)
The lengths of $40$ leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table: Find the median length of the leaves.
Show SolutionHide Solution↓
Correct Table Median class = $144.5 - 153.5$ Median = $144.5 + \frac{20-17}{12} \times 9$ $= 146.75$ Hence, median length is $146.75$ mm
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 $2$ $12$ $15$ $25$ $18$ $12$ $13$ $3$
Show SolutionHide Solution↓
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.
The following table shows the number of patients of different age group who were discharged from the hospital in a particular month : Find the 'mean' and the 'mode' of the above data.
Show SolutionHide Solution↓
Age (in years) Number of patients ($f_i$) Class Mark ($x_i$) $f_i x_i$ $5-15$ $6$ $10$ $60$ $15-25$ $11$ $20$ $220$ $25-35$ $21$ $30$ $630$ $35-45$ $23$ $40$ $920$ $45-55$ $14$ $50$ $700$ $55-65$ $5$ $60$ $300$ Total $\Sigma f_i =80$ $\Sigma f_i x_i =2830$ $1\frac{1}{2}$ for correct table Mean = $\frac{2830}{80}$ $= \frac{283}{8}$ or $35.38$ years Modal class is $35 - 45$ Mode = $35 + \frac{23-21}{2\times23-21-14} \times 10$ $= \frac{405}{11}$ or $36.82$ years (approx.)
The following table shows the number of traffic challans issued in the month of April by the traffic police : Find the 'mean' and 'mode' of the above data.
Show SolutionHide Solution↓
Table: 0-10: 3, 5, 15 10-20: 5, 15, 75 20-30: 10, 25, 250 30-40: 9, 35, 315 40-50: 2, 45, 90 50-60: 1, 55, 55 Total: 30, 800 Mean $= \frac{800}{30} = \frac{80}{3}$ or $26.67$ or $27$ (approx.) Modal class is 20-30 Mode $= 20 + \frac{10 - 5}{2 \times 10 - 5 - 9} \times 10 = \frac{85}{3}$ or $28.3$ or $28$ (approx.)
Medical check-up was carried out for 35 students of a class and their weights were recorded as follows: Find the difference between the mean weight and the median weight.
Show SolutionHide Solution↓
Correct table Mean $= 45 + \frac{14}{35} \times 2 = 45.8$ $\therefore$ Mean weight is 45.8 kg Median Class is 46 - 48 Median $= 46 + \frac{\frac{35}{2} - 14}{14} \times 2 = 46.5$ $\therefore$ Median weight is 46.5 kg Difference of mean weight and median weight $= 46.5 - 45.8 = 0.7$ kg
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.
Show SolutionHide Solution↓
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
The following table gives the daily income of $50$ cab drivers of a particular city : Income (₹): $500-600, 600-700, 700-800, 800-900, 900-1000$. No. of Drivers: $12, 14, 8, 6, 10$. Find the mean income and the modal income.
Show SolutionHide Solution↓
Correct table Mean = $750 + \frac{(-12)}{50} \times 100 = 726$ $\therefore$ Mean income is ₹ $726$ Modal Class is $600-700$ Mode = $600 + \frac{14 - 12}{(2 \times 14 - 12 - 8)} \times 100 = 625$ $\therefore$ Modal income is ₹ $625$.
Following data shows the number of family members living in different bungalows of a locality: If the median number of members is found to be $5$, find the values of $p$ and $q$.
Show SolutionHide Solution↓
Correct table. $75 + p + q = 120 \Rightarrow p + q = 45$. Median is $5 \therefore$ Median class is $4 - 6$. $5 = 4 + [\frac{120/2 - (10+p)}{60}] \times 2$. On solving, we get $p = 20$ and $q = 25$.
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'.
Show SolutionHide Solution↓
Correct table Mean $= 211$ $225 + \frac{(-7)}{13+y} \times 50 =211$ $\Rightarrow y = 12$
One healthcare center working for the welfare of the patients suffering from 'Dengue', recorded the following information : Age of Patients quad Number of Patients $0-15$ quad $8$ $15-30$ quad $5$ $30-45$ quad $x$ $45-60$ quad $16$ $60-75$ quad $12$ $75-90$ quad $9$ If the modal age of the patients is $54$, then find the value of $x$.
Show SolutionHide Solution↓
Modal class is $45 - 60$ Mode = $54$ $\therefore 45 + \left(\frac{16-x}{2\times 16-x-12}\right) \times 15 = 54$ $\Rightarrow x = 10$
Weekly expenditure on Ayurvedic medicines of few households in a locality is recorded below. Weekly Expenditure (in ₹) quad Number of Households $100-150$ quad $4$ $150-200$ quad $5$ $200-250$ quad $y$ $250-300$ quad $2$ $300-350$ quad $2$ If the mean expenditure for this is ₹$211$, then find the value of the missing frequency 'y'.
Show SolutionHide Solution↓
Correct table Mean = $211$ $\Rightarrow 225 + \frac{(-7)}{13+y} \times 50 = 211$ $\Rightarrow y = 12$
Weekly expenditure on Ayurvedic medicines of few households in a locality is recorded below. Weekly Expenditure (in ₹) Number of Households 100 - 150 4 150 - 200 5 200 - 250 y 250 - 300 2 300 - 350 2 If the mean expenditure for this is ₹211, then find the value of the missing frequency 'y'.
Show SolutionHide Solution↓
Correct table Mean = 211 $225 + \frac{(-7)}{13+y} \times 50 = 211$ $\Rightarrow y = 12$
The mode of the following frequency distribution is $55$. Find the missing frequencies 'a' and 'b'. Class Interval $0-15$ $15-30$ $30-45$ $45-60$ $60-75$ $75-90$ Total Frequency $6$ $7$ $a$ $15$ $10$ $b$ $51$
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.
Show SolutionHide Solution↓
(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
Show SolutionHide Solution↓
(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
$250$ apples of a box were weighed and the distribution of masses of the apples is given in the following table : Mass (in grams) quad $80-100$ quad $100-120$ quad $120-140$ quad $140-160$ quad $160-180$ Number of apples quad $20$ quad $60$ quad $70$ quad $x$ quad $60$ (i) Find the value of $x$ and the mean mass of the apples. (ii) Find the modal mass of the apples
Show SolutionHide Solution↓
(i)$20 + 60 + 70 + x + 60 = 250$ $x = 250-210 = 40$ 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 $10-20$ $15$ $20-30$ $x$ $30-40$ $12$ $40-50$ $20$ $50-60$ $y$ $60-70$ $8$ $70-80$ $10$
Show SolutionHide Solution↓
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 $0-50$ $4$ $50-100$ $3$ $100-150$ $5$ $150-200$ $x$ $200-250$ $y$ $250-300$ $3$ $300-350$ $4$
Show SolutionHide Solution↓
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, $19 + x + y = 30$ $\Rightarrow x = 6$
In the following table, the median age of $200$ spectators of a football match is $32$. Find the missing frequencies $p$ and $q$. Age (in years) Number of Spectators $0-10$ $20$ $10-20$ $p$ $20-30$ $50$ $30-40$ $60$ $40-50$ $32$ $50-60$ $q$
Show SolutionHide Solution↓
Age (in years) Number of consumers Cumulative frequency $0-10$ $20$ $20$ $10-20$ $p$ $20 + p$ $20-30$ $50$ $70+ p$ $30-40$ $60$ $130 + P$ $40-50$ $32$ $162 + p$ $50-60$ $q$ $162 + p + q$ Total $162 + p + q$ $1$ mark for correct table $162 + p + q = 200$ $\Rightarrow p + q = 38$ (i) Median $= 32 \Rightarrow 30-40$ is the median class. $\Rightarrow 32 = 30 + \frac{(100-70-p)}{60} \times 10$ Solving, we get $p = 18$ From (i), $q = 20$
The following distribution shows the daily pocket allowance of children of a locality. The mean daily pocket allowance is ₹36.10. Find the missing frequency, $f$. Daily pocket allowance (in $\text{Rs}$) Number of children $20-25$ $7$ $25-30$ $6$ $30-35$ $9$ $35-40$ $13$ $40-45$ $f$ $45-50$ $5$ $50-55$ $4$
The median of the following distribution is $545$. If the sum of all frequencies is $100$, then find the values of $x$ and $y$. Class quad Frequency $0-100$ quad $3$ $100-200$ quad $4$ $200-300$ quad $5$ $300-400$ quad $x$ $400-500$ quad $17$ $500-600$ quad $20$ $600-700$ quad $19$ $700-800$ quad $y$ $800-900$ quad $8$ $900-1000$ quad $3$
Show SolutionHide Solution↓
Correct table Therefore, $79 + x + y = 100$ $\Rightarrow x + y = 21$ Median class is $500 - 600$. Median = $545$ $\therefore 500 + \frac{\frac{100}{2} - (29+x)}{20} \times 100 = 545$ $\Rightarrow x = 12$ and $y = 9$
The median of 80 observations given in the following table is 138. Find the values of 'a' and 'b'. Class Interval Frequency 65-85 5 85-105 a 105-125 13 125-145 20 145-165 b 165-185 10 185-205 7
Show SolutionHide Solution↓
Correct table $55 + a + b = 80$ $\Rightarrow a + b = 25$ Median class is 125 – 145 Median = 138 $\therefore 125 + \left(\frac{80}{2} - (18+a)\right) \times \frac{20}{20} = 138$ $\Rightarrow a = 9$ and $b = 16$
Find the values of the missing frequencies $p$ and $q$ in the following distribution of $100$ observations. The median of the distribution is given as $47$.
Show SolutionHide Solution↓
Correct table $\therefore 69 + p + q = 100 \Rightarrow p+q=31$ Median class is $45 - 50$ Median $= 47$ $\therefore 45 + \left(\frac{100/2 - (29+p)}{20}\right) \times 5 = 47$ $\Rightarrow p = 13$ and $q = 18$
The mean of the following frequency distribution is $62.8$. Determine the values of $f_1$ and $f_2$. Class: $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$ Total Frequency: $5$ $f_1$ $10$ $f_2$ $7$ $8$ $50$
Show SolutionHide Solution↓
Correct table Mean = $62.8$ $\frac{2060 + 30f_1 + 70f_2}{50} = 62.8$ $\Rightarrow 30f_1+70f_2 = 1080$ or $3f_1 + 7f_2 = 108$ --- (1) Also, $30 + f_1 + f_2 = 50$ $\Rightarrow f_1 + f_2 = 20$ --- (2) Solving (1) and (2), we get $f_1 = 8$ and $f_2 = 12$
The median of the following frequency distribution is $35$. Find the value of $x$ and hence find the mode. Class: $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ Frequency: $2$ $3$ $x$ $6$ $5$ $3$ $2$
Show SolutionHide Solution↓
Correct table Median class is $30 – 40$ Median = $35$ $\therefore 30 + \frac{(\frac{21+x}{2})-(5+x)}{6} \times 10 = 35$ $\Rightarrow x = 5$ Modal class is $30 – 40$ Mode = $30+\frac{6-5}{2\times6-5-5} \times 10$ $= 35$
Following distribution shows the marks of 230 students in a particular subject. If the median marks are 46, then find the values of $x$ and $y$.
Show SolutionHide Solution↓
Correct table (1 mark). $150 + x + y = 230 \Rightarrow x + y = 80$ (1 mark). Median is 46 $\therefore$ Median class is 40 - 50 ($\frac{1}{2}$ mark). $46 = 40 + [\frac{230/2 - (42+x)}{65}] \times 10$ (1 mark). On solving, we get $x = 34$ and $y = 46$ (1 + $\frac{1}{2}$ marks).
The population of lions was noted in different regions across the world in the following table : If the median of the given data is $525$, find the values of $x$ and $y$.
Show SolutionHide Solution↓
Correct table $74 + x + y = 100 \implies x + y = 26$ Median class is $500 - 600$ $525 = 500 + [\frac{50 - (28+x)}{20}] \times 100$ On solving, we get $x = 17, y = 9$
This section has 20 Multiple Choice Questions (MCQs) carrying 1 mark each. What is the mode of a data if median and mean of the same data are $9-6$ and $10-5$, respectively?
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.
Show SolutionHide Solution↓
Sol. 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.
Show SolutionHide Solution↓
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 $\frac{3}{5}$
_q_1.png)
_a_1.png)
_a_1.png)
_a_1.png)
