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Maths 3 Marks Question

Data Handling · CBSE STD 7 (CBSE - English Medium). 8 questions.

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3 Marks Question

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8 questions · timed · auto-graded

Question 13 Marks
Following cards are put facing down:

What is the chance of drawing out
  1. a vowel
  2. a card marked $U$
  3. $A$ or $I$
  4. a consonant
Answer
  1. We can clearly see that all the $5$ letters are vowels, i.e. $A, E, I, O,$ Hence, it is certain to draw a vowel, i.e. probability of drawing a vowel is $1.$
  2. Probability $=\frac{\text{Number of cards marked U}}{\text{Total number of cards}}=\frac{1}{5}$
  3. Probability $=\frac{\text{Number of cards marked A or I}}{\text{Total number of cards}}=\frac{2}{5}$
  4. Probability $=\frac{\text{Number of cards marked with a constant}}{\text{Total number of cards}}=\frac{0}{5}=0$
Hence, it is impossible to draw a constant.
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Question 23 Marks
Age (in years) of $6$ children of two groups are recorded as below:
Age (in Years)
Group A
Group B
$7$
$7$
$7$
$9$
$9$
$11$
$8$
$12$
$10$
$12$
$10$
$12$
  1. Find the mode and range for each group.
  2. Find the range and mode if the two groups are combined together.
Answer
From the given table, age of children in group $A = 7$ year, $7$ year, $9$ year, $8$ year, $10$ year, $10$ year
Age of children in group $8 = 7$ year, $9$ year, $11$ year, $12$ year, $12$ year, $12$ year
  1. Mode in group $A = 7$ year and $10$ year. $[\because 7$ year and $10$ year occurs most frequent, i.e. $2$ times$]$
Range in group $A =$ Maximum value $-$ Minimum value $= 10 - 7 = 3$
Mode in groups $= 12$ year $[\because 12$ year is the most frequent, i.e. $3]$
Range in group $8 =$ Maximum value $-$ Minimum value $= 12 - 7 = 5$
  1. If both groups are combined together $7, 7, 7, 9, 9, 11, 8, 12, 10, 12, 10, 12.$
Mode $= 7$ and $12 [\because 7$ and $12$ occurs most frequent, i.e. $3$ times$]$
$\therefore$ Range $=$ Maximum value $-$ Minimum value $= 12 - 7 = 5$
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Question 33 Marks
For the given data below, calculate the mean of its median and mode
$6, 2, 5, 4, 3, 4, 4, 2, 3$
Answer
Given, data in ascending order is $2, 2, 3, 4, 4, 4, 5, 6$
Hence, $n = 9($odd$)$
Median $=$ Value of $\Big(\frac{\text{n+1}}{2}\Big)\text{th}$
observation $=\Big(\frac{9+1}{2}\Big)\text{th}$
observation $= 5th$ observation $= 4$
Mode $=$ Most frequant observation $= 4$
Mean is the central value.
$\therefore\ \text{Mean}=\frac{1}{2}[3\times\text{median}-\text{mode}]$
$=\frac{1}{2}[3\times4-4]=\frac{1}{2}[12-4]$
$=\frac{1}{2}\times8=4$
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Question 43 Marks
A die was thrown $15$ times and the outcomes recorded were $5, 3, 4, 1, 2, 6, 4, 2, 2, 3, 1, 5, 6, 1, 2.$ Find the mean, median and mode of the data.
Answer
Given data is $5, 3, 4,1,2, 6, 4, 2, 2, 3,1,5, 6,1,2$
Arranging the data in ascending order, we have $1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6.$
$\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$=\frac{1+1+1+2+2+2+2+3+3+4+4+5+5+6+6}{15}$
$=\frac{47}{15}=3.13$
Mode $=$ Most frequent data $= 2$
Here, $n = 15($odd$)$
Median $=$ Value of $\Big(\frac{\text{n}+1}{2}\big)\text{th }$
observation $=$ Value of $\Big(\frac{15+1}{2}\Big)\text{th}$
observation $=$ Value of $8th$ observation $= 3$
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Question 53 Marks
A car seller collects the following data of cars sold in his shop.
Colour of Car
Number of Cars Sold
Red
$15$
Black
$20$
White
$17$
Silver
$12$
Others
$9$
 
  1. Which colour of the car is most liked$?$
  2. Which measure of central tendency was used in $(a)?$
Answer
  1. Red colour of the car liked by people $= 15$
Black colour of the car liked by people $= 20$
White colour of the car liked by people $= 17$
Silver colour of the car liked by people $= 12$
Other colour of the car liked by people $= 9$
Hence, black colour of the car is the most liked.
  1. Mode concept used in $(a).$
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Question 63 Marks
Calculate the mean, median and mode of the following data: $5, 10, 10, 12, 13 $ Are these three equal$?$
Answer
Given data is $5, 10, 10, 12, 13.$
Sum of all observations $= 5 + 10 + 10 + 12 + 13 = 50$
Number of observations $= 5$
$\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total observations}}=\frac{50}{5}=10$
Here, $n = 5($odd$)$
So, median $=$ value of $\Big(\frac{\text{n+1}}{2}\Big)\text{th}$
observation $=$ value of $\Big(\frac{5+1}{2}\Big)\text{th}$
observation $=$ value of $3rd$
observation $= 10 $ Mode $=$ Most frequant data $= 10$
Hence, Mean $=$ Median $=$ Mode Yes,
Mean, median and mode are equal.
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Question 73 Marks
The mean of three numbers is $10.$ The mean of other four numbers is $12.$ Find the mean of all the numbers.
Answer
Mean of three numbers $=\frac{\text{Sum of threee numbers}}{3}$
$\Rightarrow10=\frac{\text{Sum of three numbers}}{3}$ $[\because$ mean of three numbers $= 10,$ given$]$
Hence, sum of three numbers $= 30$
Mean of other four numbers $=\frac{\text{Sum of other four numbers}}{4}$
$\Rightarrow12=\frac{\text{Sum of all the numbers}}{\text{Total numbers}}$
$[\because$ mean of other four numbers $= 12$, given$]$
Hence, sum of other four numbers $= 48$
$\therefore$ Mean of all the numbers $=\frac{\text{Sum of threee numbers}}{\text{Total numbers}}$
$=\frac{\text{[Sum of first three numbers $+$ Sum of other four numbers]}}{7}$
$=\frac{30+48}{7}=\frac{78}{7}=11.14$
Hence, mean of all the numbers is $11.14.$
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Question 83 Marks
Find the median of the given data if the mean is $4.5. 5, 7, 7, 8, x, 5, 4, 3, 1, 2$
Answer
Given, mean $= 4.5$ We know that,
$\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$\Rightarrow4.5=\frac{5+7+7+8+\text{x}+5+4+3+1+2}{10}$
$\Rightarrow4.5\times10=42+\text{x}$
$\Rightarrow45-12=\text{x}$
$\therefore\ \text{x}=3$
Now, arrange the data in ascending order $1, 2, 3, 3, 4, 5, 5, 7, 7, 8$
Here $n = 10($even$)$
$\text{Median}=\frac{\text{Value of}\Big(\frac{\text{n}}{2}\Big)\text{th observation}+\text{Value of}\Big(\frac{\text{n}}{2}+1\Big)\text{th observation}}{2}$
$=\frac{\text{Value of}\ 5\text{th observation}+\text{Value of} \ 6\text{th observation}}{2}$
$=\frac{4+5}{2}=\frac{9}{2}=4.5$
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