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MATHS 5 Marks Questions

Data Handling II (Central Values) · UP Board (UPMSP) STD 7 (UPMSP - English Medium). 20 questions.

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Question 15 Marks
The mean of marks scored by $100$ students was found to be $40$. Later on it was discovered that a score of $53$ was misread as $83.$ Find the correct mean.
Answer
We have, $N =$ The number of observations $= 100,$
Mean $= 40$
$\text{Mean}=\frac{\text{Sum of the observations}}{\text{Total number of observations}}$
$\Rightarrow40=\frac{\text{Sum of the observations}}{100}$ Sum of the observations $= 40 \times 100$
Thus, the incorrect sum of the observations $= 40 \times 100 = 4000$
Now, The correct sum of the observations = Incorrect sum of the observations – Incorrect observation + Correct observation.
The correct sum of the observations $= 4000 - 83 + 53$
The correct sum of the observations $= 4000 - 30 = 3970$
$\therefore\text{Correct mean}=\frac{\text{Correct sum of the observations}}{\text{Number of observations}}=\frac{3970}{100}=39.7$
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Question 25 Marks
Heights of 25 children (in cm) in a school are as given below: $168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 164, 163, 160, 165, 163, 162$ What is the mode of heights? Also, find the mean and median.
Answer
Arranging the data in tabular form, we get:

Here, $\mathrm{n}=25$
Median $=$ value of $\frac{\mathrm{n}+1}{2}$ th observation $=$ value of the $13^{\text {th }}$ observation $=163 \mathrm{~cm}$.
Here, clearly, 163 cm occurs the most number of times. Therefore, the mode of the given data is $163 cm .$
Mode $=3$ Median -2 Mean
$\Rightarrow 163=3 \times 163-2 \text { Mean }$
$\Rightarrow 2 \text { Mean }=326$
$\Rightarrow \text { Mean }=163 \mathrm{~cm}$
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Question 35 Marks
The mean of the following data is $20.6.$ Find the value of $p.$
x:
 $10$
$15$
$p$
$25$
 $35$
f:
 $3$
$10$
$25$
$7$
 $5$
Answer
Calculation of mean:
$x_i$
$f_i$
$x_if_i$
$10$
$3$
$30$
$15$
$10$
$150$
$p$
$25$
$25p$
$25$
$7$
$175$
$35$
$5$
$175$
Total
$\sum\text{f}_\text{i}=50$
$\sum\text{f}_\text{i}\text{x}_\text{i}=530\ +\ 25\text{p}$
We have,
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow20.6=\frac{530+25\text{p}}{50}$
$\Rightarrow530 + 25\text{p} = 20.6 × 50$
$\Rightarrow25\text{p} = 1030 - 530$
$\Rightarrow\text{p}=\frac{500}{25}$
$\Rightarrow\text{p} = 20$
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Question 45 Marks
If the mean of the following data is $15$, find $p$.
x:
 $5$
$10$
$15$
$20$
 $25$
f:
 $6$
$p$
$6$
$10$
 $5$
Answer
Calculation of mean:
$x_i$
$f_i$
$x_if_i$
$5$
$6$
$30$
$10$
$p$
$10p$
$15$
$6$
$90$
$20$
$10$
$200$
$25$
$5$
$125$
Total
$\sum\text{f}_\text{i}=27+\text{p}$
$\sum\text{f}_\text{i}\text{x}_\text{i}=445+10\text{p}$
We have,
$\sum\text{f}_\text{i}=27+\text{p},\sum\text{f}_\text{i}\text{x}_\text{i}=445+10\text{p}$
$\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow20.6=\frac{445+10\text{p}}{27+\text{p}}$
$\Rightarrow445 + 10\text{p} = 405 + 15\text{p}$
$\Rightarrow5\text{p} = 445 - 405$
$\Rightarrow\text{p}=\frac{40}{5}$
$\Rightarrow\text{p} = 8$
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Question 55 Marks
The mean of $200$ items was $50$. Later on, it was discovered that the two items were misread as $92$ and $8$ instead of $192$ and $88$. Find the correct mean.
Answer
N = Number of observations $= 200$
$\text{Mean}=\frac{\text{Sum of the observations}}{\text{Number of observations}}$
$\Rightarrow50=\frac{\text{Sum of the observations}}{200}$
$\Rightarrow $ Sum of the observations $= 50 \times 200 = 10,000.$
Thus, the incorrect sum of the observations $= 50 \times 200$
Now, The correct sum of the observations = Incorrect sum of the observations - Incorrect observations + Correct observations.
$\Rightarrow $ Correct sum of the observations $= 10,000 - (92 + 8) + (192 + 88)$
$\Rightarrow $ Correct sum of the observations $= 10,000 - 100 + 280$
$\Rightarrow $ Correct sum of the observations $= 9900 + 280$
$\Rightarrow $ Correct sum of the observations $= 10,180$
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Question 65 Marks
Find the missing frequency $(p)$ for the following distribution whose mean is $7.68$
x:
 $3$
$5$
$7$
$9$
 $11$ $13$
f:
 $6$
$8$
$15$
$p$
 $8$ $4$
Answer
Calculation of mean:
$x_i$
$f_i$
$x_if_i$
$3$
$6$
$18$
$5$
$8$
$40$
$7$
$15$
 $105$
$9$
$p$
 $9p$
$11$ $8$ $88$
$13$ $4$ $52$
Total
$\sum\text{f}_\text{i}=41+\text{p}$
$\sum\text{f}_\text{i}\text{x}_\text{i}=303+9\text{p}$
We have,
$\sum\text{f}_\text{i}=41+\text{p},\sum\text{f}_\text{i}\text{x}_\text{i}=303+9\text{p}$ $\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow7.68=\frac{303+9\text{p}}{41+\text{p}}$
$\Rightarrow303 + 9\text{p} = 314.88 + 7.68\text{p}$
$\Rightarrow1.32\text{p} = 314.88 - 303$
$\Rightarrow\text{p}=\frac{11.88}{1.32}$
$\Rightarrow\text{p} = 9$
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Question 75 Marks
Find the value of $p$ for the following distribution whose mean is $16.6$
x:
 $8$
$12$
$15$
$p$
 $20$ $25$ $30$
f:
 $12$
$16$
$20$
$24$
 $16$ $8$ $4$
Answer
Calculation of mean:
$x_i$
$f_i$
$x_if_i$
$8$
$12$
$96$
$12$
$16$
$192$
$15$
$20$
$300$
$p$
$24$
$24p$
$20$
$16$
$320$
$25$ $8$ $200$
$30$ $4$ $120$
Total
$\sum\text{f}_\text{i}=\text{N}=100$
$\sum\text{f}_\text{i}\text{x}_\text{i}=1288+24\text{p}$
We have,
$\sum\text{f}_\text{i}=100+\text{p},\sum\text{f}_\text{i}\text{x}_\text{i}=1228+24\text{p}$ $\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow16.6=\frac{1228+24\text{p}}{100}$
$\Rightarrow1228 + 24\text{p} = 16.6 × 100$
$\Rightarrow24\text{p} = 1660 - 1228$
$\Rightarrow\text{p} = 18$
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Question 85 Marks
Find the missing value of p for the following distribution whose mean is $12.58$
x:
 $5$ $8$
$10$
$12$
 $p$ $20$ $25$
f:
 $2$
$5$
$8$
 $22$ $7$ $4$ $2$
Answer
Calculation of mean:
$x_i$
$f_i$
$x_if_i$
$5$
$2$
$10$
$8$
$5$
$40$
$10$
$8$
$80$
$12$
$22$
$264$
$p$
$7$
$7p$
$20$ $4$ $80$
$25$ $2$ $50$
Total
$\sum\text{f}_\text{i}=\text{N}=50$
$\sum\text{f}_\text{i}\text{x}_\text{i}=524+7\text{p}$
We have,
$\sum\text{f}_\text{i}=50,\sum\text{f}_\text{i}\text{x}_\text{i}=524+7\text{p}$ $\therefore\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow12.58=\frac{524+7\text{p}}{50}$
$\Rightarrow524 + 7\text{p} = 12.58 × 50$
$\Rightarrow7\text{p} = 629 - 524$
$\Rightarrow\text{p}=\frac{105}{7}$
$\Rightarrow\text{p} = 15$
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Question 95 Marks
Find the value of p, if the mean of the following distribution is $20$
x:
 $15$
$17$
$19$
$20 + p$
 $23$
f:
 $2$
$3$
$4$
$5 p$
 $6$
Answer
Calculation of mean:
$x_i$
$f_i$
$x_if_i$
$15$
$2$
$30$
$17$
$3$
$51$
$19$
$4$
 $76$
$20 + p$
$5p$
 $(20 + p) 5p$
$23$ $6$ $138$
Total
$\sum\text{f}_\text{i}=15+5\text{p}$
$\sum\text{f}_\text{i}\text{x}_\text{i}=295+(20+\text{p})\ 5\text{p}$
We have,
$\sum \mathrm{f}_{\mathrm{i}}=15+5 \mathrm{p}, \sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=295+(20+\mathrm{p}) 5 \mathrm{p}$
$\therefore \text { Mean }=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$
$\Rightarrow 20=\frac{(295+(20+\mathrm{p}) 5 \mathrm{p})}{15+5 \mathrm{p}}$
$\Rightarrow 295+100 \mathrm{p}+5 \mathrm{p}^2=300+100 \mathrm{p}$
$\Rightarrow 5 \mathrm{p}^2=300-295$
$\Rightarrow 5 \mathrm{p}^2=5$
$\Rightarrow \mathrm{p}^2=1$
$\Rightarrow \mathrm{p}=1$
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Question 105 Marks
The following table shows the weights of $12$ persons.
Weight (in kg): $48$ $50$ $52$ $54$ $58$
Number of persons: $4$ $3$ $2$ $2$ $1$
Find the median and mean weights. Using empirical relation, calculate its mode.
Answer
Weight ($x_i$) $48$ 50 52 54 58 Total
Number of Persons ($f_i$) $4$ $3$ $2$ $2$ $1$ $\sum\text{f}_\text{i}=12$
($f_ix_i$) $192$ $150$ $104$ $108$ $58$ $\sum\text{f}_\text{i}\text{x}_\text{i}=612$
$\text{mean} = \frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}\frac{612}{12}=51\text{kg.}$
Here, $n = 12$
Median = value of $\frac{\text{n}}{2}\text{th}$ observation $\frac{\text{n}}{2}+1^\text{th}$ observation
$\Rightarrow\text{Median}=\frac{\text{Value of 6}^\text{th}\ \text{observation}+\text{Value of 7}^\text{th}\ \text{observation}}{2}$
$\Rightarrow\text{Median}=\frac{50+50}{2}=50\text{kg.}$
Now,
Mode $= 3$ Median $- 2$ Mean
$\Rightarrow Mode = 3 \times 50 - 2 \times 51$
$\Rightarrow Mode = 150 - 102$
$\Rightarrow Mode = 48kg.$
Thus, Mean $= 51kg$, Median $= 50 kg$ and Mode $= 48kg.$
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Question 115 Marks
The following table shows the weights of 12 persons.
Weight (in kg):4850525458
Number of persons:43221
Find the median and mean weights. Using empirical relation, calculate its mode.
Answer
Mean = 51 kg, Mode = 48 kg, Median = 50 kg.
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Question 125 Marks
Calculate the mean and median for the following data:
Marks:1011121314161920
Number of students:35452321
Using empirical formula, find its mode.
Answer
Mean = 13.28, Mode = 12.44, Median = 13
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Question 135 Marks
Heights of 25 children (in cm) in a school are as given below:
168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 165, 163, 162, 163, 164, 163, 160, 165, 163, 162. What is the mode of heights? Also, find the mean and median.
Answer
Mode = 163 cm = Mean = Median
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Question 145 Marks
Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14. By using the empirical relation also find the mean.
Answer
Mode = 14, Median = 14, Mean = 14
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Question 155 Marks
Find the value of p for the following distribution whose mean is 16.6
x:81215p202530
f:121620241684
Answer
18
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Question 165 Marks
The daily wages (in ₹) of 15 workers in a factory are given below:
200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180
Prepare the frequency table and find the mean wage.
Answer
Wages (xi):130150180200
No. of workers (fi):2463
Mean wage = ₹ 169.33
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Question 185 Marks
Find the missing frequency (p) for the following distribution whose mean is 7.68
x:35791113
f:6815p84
Answer
9
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Question 195 Marks
Find the missing value of p for the following distribution whose mean is 12.58
x:581012p2025
f:25822742
Answer
15
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Question 205 Marks
A die was thrown 20 times and the following scores were recorded:
5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1
Prepare the frequency table of the scores on the upper face of the die and find the mean score.
Answer
Score (xi):123456
Frequency (fi):251462
Mean score = 3.65
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