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MATHS M.C.Q

Statistics and Probability · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 30 questions.

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M.C.Q

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30 questions · timed · auto-graded

Question 11 Mark
Write the correct answer in the following:
In a medical examination of students of a class, the following blood groups are recorded:
Blood group
A
AB
B
O
Number of Student
10
13
12
5
A student is selected at random from the class. The probability that he/ she has blood group B, is:
  1. $\frac{1}{4}$
  2. $\frac{13}{40}$
  3. $\frac{3}{10}$
  4. $\frac{1}{8}$
Answer
  1. $\frac{3}{10}$
Solution:
Total number of students = 10 + 13 + 12 + 5 = 40
Let us call event of selected students at random who have blood group B be E.
There are 12 students who has blood group B.
So, $\text{P(E)}=\frac{12}{40}=\frac{3}{10}$
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MCQ 21 Mark
Write the correct answer in the following: If $\bar{\text{x}}$ represents the mean of n observations $x_1,_{ }x_2, ... x_n,$ then value of $\sum\limits_{\text{i}=1}^\text {b} \text{x}_\text{i}-\bar{\text{x}}$ is:
  • A
    $-1$
  • ✓
    $0$
  • C
    $1$
  • D
    $n - 1$
Answer
Correct option: B.
$0$
We know that algebraic sun of deviations from mean is zero.
$\sum\limits_{\text{a}=1}^\text{b} (\text{x}_\text{t}-\bar{\text{x}})$
$=(\text{x}_1-\bar{\text{x}})+(\text{x}_2-\bar{\text{x}})+(\text{x}_3-\bar{\text{x}})+\ ...\ +(\text{x}_\text{n}-\bar{\text{x}})$
$= (\text{x}_1+\text{x}_2+\text{x}_3+... +\text{x}_\text{n})- \text{n}\bar{\text{x}}$
$\Rightarrow\sum\limits_{\text{t}-1}^\text{b}\text{x}_\text{i}-\text{n}\bar{\text{x}}=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}=0$ $\bigg[\because\sum\limits_{\text{i}=1}^\text{n} \text{x}_\text{i}=\text{n}\bar{\text{x}}\bigg]$
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Question 31 Mark
Write the correct answer in the following:
A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data:
268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236.
The frequency of the class 310-330 is:
  1. 4
  2. 5
  3. 6
  4. 7
Answer
  1. 6
Solution:
The observation corresponding to class 310–330 (330 not included in this interval) are 310, 310, 320, 319, 318, 316, i.e., 6 observations.
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Question 41 Mark
Write the correct answer in the following:
The range of the data: 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is:
  1. 10
  2. 15
  3. 18
  4. 26
Answer
  1. 26
Solution:
Maximum value of the variate = 32
And the minimum value of the variate = 6
Range = Maximum value of the variate-Minimum value of the variate = 32 - 6 = 26
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Question 51 Mark
Write the correct answer in the following:
To draw a histogram to represent the following frequency distribution:
Class intervai
5-10
10-15
15-25
25-45
45-75
Frequency
6
12
10
8
15
The adjusted frequency for the class 25-45 is:
  1. 6
  2. 5
  3. 3
  4. 2
Answer
  1. 2
Solution:
The adjusted frequency for the class 25 - 45 is
$= \frac{\text{Frequency of the class }}{\text{Class width }}\times \text{Minimum width }=\frac{8}{20} \times5=2$
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Question 61 Mark
Write the correct answer in the following:
The probability that bulbs selected randomly from the lot has life less than 900 hours is:
  1. $\frac{11}{40}$
  2. $\frac{5}{16}$
  3. $\frac{7}{16}$
  4. $\frac{9}{16}$
Answer
  1. $\frac{9}{16}$
Solution:
Total number of bulbs in a lot, n(S) = 80
Number of bulbs whose life time is less than 900h, n(E) = 10 + 12 + 23 = 45
Probability that bulbs has life time less than 900h $=\frac{\text{n(E)}}{\text{n(S)}}=\frac{45}{80}=\frac{9}{16}$
Hence, the probability that bulb has life time less than 900 is $\frac{9}{16}$
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Question 71 Mark
Write the correct answer in the following:
The width of five continuous classes in a frequency distribution is 5 and the lower class-limit of the lowest class is 10. The upper class-limit of the highest class is:
  1. 15
  2. 25
  3. 35
  4. 40
Answer
  1. 35
Solution:
Sol. Width of each of the five continuous classes in a frequency distribution is 5.
Lower class limit of the lowest class = 10
Upper class limit of the lowest class is 10 + 5 = 15
So, the five continuous classes are
10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35
Hence, the upper-class limit of the height class is 35.
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Question 81 Mark
Write the correct answer in the following:In the class intervals 10–20, 20–30, the number 20 is included in:
  1. 10–20
  2. 20–30
  3. Both the intervals.
  4. None of these intervals.
Answer
  1. 20–30
Solution:
The number 20 is included in 20–30.
Hence, (b) is the correct answer.
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Question 91 Mark
Write the correct answer in the following:
A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data:
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44. The number of classes in the distribution will be:
  1. 9
  2. 10
  3. 11
  4. 12
Answer
  1. 10
Solution:
Minimum value = 14
Maximum value = 112
The classes are
13 - 22, 23 - 32, 33 - 42, 43 - 52, 53 - 62, 63 - 72, 73 - 82, 83 - 92, 93 - 102 and 103 - 112.
The number of classes in the distribution will be 10.
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MCQ 101 Mark
Let $\bar{\text{x}}$ be thae mean of $x_1, x_2, ......, x_n,$ and $\bar{\text{y}}$ the mean of $y_1 , y_2 , ... , y_n$ . If $\bar{\text{z}}$ is the mean of $x_1 , x_2 , ... , x_n , y_1 , y_2, ... , yn,$ then $\bar{\text{z}}$ is equal to:
  • A
    $\bar{\text{x}}+\bar{\text{y}}$
  • ✓
    $\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$
  • C
    $\frac{\bar{\text{x}}+\bar{\text{y}}}{\text{n}}$
  • D
    $\frac{\bar{\text{x}}+\bar{\text{y}}}{2\text{n}}$
Answer
Correct option: B.
$\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$
We have $\bar{\text{x}}$ is the mean of $x_1, x_2, ......, x_n,$ and $\bar{\text{y}}$ is the mean of $y_1 , y_2 , ... , y_n$
So, $\bar{\text{x}}=\frac{(\text{x}_1+\text{x}_2+\text{x}_3+\ ....\ +\text{x}_\text{n})}{\text{n}}$
$\Rightarrow\text{x}_1+\text{x}_2+\text{x}_3+\ .....\ +\text{x}_\text{n}=\text{n}\bar{\text{x}}$
And $\bar{\text{y}}=\frac{(\text{y}_1+\text{y}_2+\text{y}_3+\ ....\ +\text{y}_\text{n})}{\text{n}}$
$\Rightarrow\text{y}_1+\text{y}_2+\text{y}_3+\ ....\ +\text{y}_\text{n}=\text{n}\bar{\text{y}}$
If $\bar{\text{z}}$ is the mean of $x_1, x_2, ...., y_1 , y_2 , ... , y_n$  then
{$\bar{\text{z}}=\frac{\text{n}\bar{\text{x}}+\text{n}\bar{\text{y}}}{\text{n}+\text{n}}=\frac{\text{n}(\bar{\text{x}}+\bar{\text{y}})}{2\text{n}}=\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$}
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Question 111 Mark
Write the correct answer in the following:
The class marks of a frequency distribution are given as follows: 15, 20, 25, .... The class corresponding to the class mark 20 is:
  1. 12.5 - 17.5
  2. 17.5 - 22.5
  3. 18.5 - 21.5
  4. 19.5 - 20.5
Answer
  1. 17.5 - 22.5
Solution:
The class mark are 15, 20, 25, ….
The size of each class interval is 25 - 20 = 20 - 15 = 5
Hence, the class interval corresponding to the class mark 20 is,
(20 - 2.5) - (20 + 2.5) i.e., 17.5 - 22.5.
So, (b) is the correct answer.
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Question 121 Mark
Write the correct answer in the following:
Let m be the mid-point and l be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:
  1. 2m + l
  2. 2m - l
  3. m - l
  4. m - 2l
Answer
  1. 2m - l
Solution:
Let x and y be the lower and upper class limit of a continuous frequency distribution.
Now, mid-point of a class $=\frac{(\text{x}+\text{y})}{2}=\text{m}$ [given]
⇒ x + y = 2m = x + l = 2m
$\big[\therefore$ y = l = upper class limit (given)$\big]$
⇒ x = 2m - l
Hence, the lower class limit of the class is 2m - l.
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Question 131 Mark
Write the correct answer in the following:
In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he/ she does not like to eat potato chips is:
  1. 0.25
  2. 0.50
  3. 0.75
  4. 0.80
Answer
  1. 0.75
Solution:
Total number of survey children’s age from 19 - 36 months, n(S) = 364 In those of them 91 out of them liked to eat potato chips.
$\therefore$ Number of children who do not like to eat potato chips, n(E) = 364 - 91 = 273
$\therefore$ Probability that he/she does not like to eat potato chips $=\frac{\text{n(E)}}{\text{n(s)}}=\frac{273}{364}=0.75$
Hence, the probability that he/ she does not like to eat potato chips is 0.75.
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MCQ 141 Mark
Write the correct answer in the following:
The mean of $25$ observations is $36$. Out of these observations if the mean of first $13$ observations is $32$ and that of the last $13$ observations is $40$, the $13^{th}$ observation is:
  • A
    $23$
  • ✓
    $36$
  • C
    $38$
  • D
    $40$
Answer
Correct option: B.
$36$
Mean of first $13$ observation $= 32$
$\therefore$ Sum of all first $13$ observation $= (32 \times 13) = 416$
Mean of last $13$ observation $= 40$
$\therefore$ Sum of all last $13$ observation $= (40 \times 13) = 520$
Mean of $25$ observation $= 36$
$\therefore$ Sum of all first $25$ observation $= (36 \times 25) = 900$
Hence$, 13^{th}$ observation $= 416 + 520 - 900 = 36$
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Question 151 Mark
Write the correct answer in the following:
Mode of the data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15 is:
  1. 14
  2. 15
  3. 16
  4. 17
Answer
  1. 15
Solution:
We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20 From above, we see that 15 occurs most frequently i.e., 5 times.
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Question 161 Mark
Write the correct answer in the following:
In a frequency distribution, the mid value of a class is 10 and the width of the class is. The lower limit of the class is:
  1. 6
  2. 7
  3. 8
  4. 12
Answer
  1. 7
Solution:
Let x and y be the uppel and lower and lower class limit in a frequency distribution.
Now, mid value of a class $\frac{(\text{x}+\text{y})}{2}=10$ [given]
⇒ x + y = 20 ...(i)
Also, given that, width of class x - y = 6 ...(ii)
On adding Eqs. (i) and (ii), we get
2x = 20 + 6
⇒ 2x = 26
⇒ x = 13
On putting x = 13 in Eq. (i), we get
13 + y = 20
⇒ y = 7
Hence, the lower limit of the class is 7.
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MCQ 171 Mark
If $\bar{\text{x}}_1, \bar{\text{x}}_2, \bar{\text{x}}_3, ....., \bar{\text{x}}_\text{n}$ are the means of $n$ group with $n_1, n_2, ...., nn$ number of observations respectively, then the mean $\bar{\text{x}}$ of all the groups taken together is given by:
  • A
    $\sum\limits^\text{n}_{\text{i}=1}$
  • ✓
    $\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\text{n}^2}$
  • C
    $\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\text{n}_2}$
  • D
    $\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{2\text{n}}$
Answer
Correct option: B.
$\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\text{n}^2}$
If $\bar{\text{x}}_1, \bar{\text{x}}_2, \bar{\text{x}}_3, ....., \bar{\text{x}}_\text{n}$ are the mean of $n$ group with $n_1, n_2, ...., n_n$ number of observation respectively, then the mean $\bar{\text{x}}$
$\bar{\text{x}}=\frac{\text{n}_1​​​​\bar{\text{x}}_1+\text{n}_2\bar{\text{x}}_2+\text{n}_3\bar{\text{x}}_3+\ ....\ +\text{n}_\text{n}\bar{\text{x}}_\text{n}}{\text{n}_1+\text{n}_2+\text{n}_3+\ ....\ +\text{n}_\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}}$
Hence, the mean $\bar{\text{x}}$ of all group taken together is gien up $=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}}$
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Question 181 Mark
The mean of 100 observations is 50. If one of the observations which was 50 is replaced by 150, the resulting mean will be:
  1. 50.5
  2. 51
  3. 51.5
  4. 52
Answer
  1. 51
Solution:
We have, $\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{n}}$
$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow\sum\text{x}_\text{i}=50\times100=5,000$
If one of the observation which was 50 is resplaced by 150, then
$\sum\text{x}_\text{i}=5,000-50+150=5100$
Then, the resulting mean $=\frac{5100}{100}=51$
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Question 191 Mark
Write the correct answer in the following:
For drawing a frequency polygon of a continous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abcissae are respectively:
  1. Upper limits of the classes.
  2. Lower limits of the classes.
  3. Class marks of the classes.
  4. Upper limits of perceeding classes.
Answer
  1. Class marks of the classes.
Solution:
Abcissac are the class marks of the classes.
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MCQ 201 Mark
Write the correct answer in the following: If each observation of the data is increased by $5,$ then their mean:
  • A
    Remains the same.
  • B
    Becomes $5$ times the original mean.
  • C
    Is decreased by $5.$
  • ✓
    Is increased by $ 5.$
Answer
Correct option: D.
Is increased by $ 5.$
Let $x_1, x_2, ...., x_n$ be the $n$ observation,
Then, old mean $\bar{\text{x}}_{\text{old}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}\ \dots(\text{i})$
Now, adding $5$ in each obsevation, the new mean becomes
$\bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+5)+(\text{x}_2+5)+\ ...\ +(\text{x}_\text{n}+5)}{\text{n}}$
$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+\text{x}_2+\ ...\ +\text{x}_\text{n})+5\text{n}}{}$
$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}+5=\bar{\text{x}}_{\text{old}}+5 \ [$from $Eq. (i)]$
$\Rightarrow\ \bar{\text{x}}_\text{new}=\bar{\text{x}}_{\text{old}}+5$
Hence, the new mean is increased by $5.$
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Question 211 Mark
Write the correct answer in the following:
In a sample study of 642 people, it was found that 514 people have a high school certificate. If a person is selected at random, the probability that the person has a high school certificate is:
  1. 0.5
  2. 0.6
  3. 0.7
  4. 0.8
Answer
  1. 0.8
Solution:
We know that empricrical probability P(E) of an event E happen, is given by
$\text{P(E)}=\frac{\text{Number of trials in which the event happed}}{\text{Total number of trials}}$
Let E be the event that a person selected at random has a high school certificate.
$\text{P(E)}=\frac{\text{Numbers of person having high school certificate}}{\text{Total number of people in a sample study}}$
$=\frac{514}{642}=0.8$
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Question 221 Mark
Write the correct answer in the following:
The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is:
  1. 45
  2. 49.5
  3. 54
  4. 56
Answer
  1. 54
Solution:
Arranging the data in ascending order, we get
22, 34, 39, 45, 54, 56, 78, 84
Here n = 9, which is an odd number.
$\therefore\ \text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}$
$\text{Value}=\Big(\frac{9+1}{2}\Big)^{\text{th}}$
$\text{value}=5^{\text{th}}\text{value}$
So, median = 54
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Question 231 Mark
Write the correct answer in the following:
Median of the following numbers, 4, 4, 5, 7, 6, 7, 7, 12, 3 is:
  1. 4
  2. 5
  3. 6
  4. 7
Answer
  1. 6
Solution:
First, we arrange the given numbers in ascending order is,
3, 4, 4, 5, 6, 7, 7, 7 and 12
Here, n = 9
Since, n is odd, so we use the formula for median,
Now, Median $=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}$
$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{observation}$ [Put n = 9]
$=\Big(\frac{10}{2}\Big)^{\text{th}}\text{observation}$
$=5^{\text{th}}\text{observation}=6$ 
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Question 241 Mark
Write the correct answer in the following:
80 bulbs are selected at random from a lot and their life time (in hrs) is recorded in the form of a frequency table given below:
Life time(in hours)
300
500
700
 900 1100
Frequency
 10
12
23
 25 10
One bulb is selected at random from the lot. The probability that its life is 1150 hours, is:
  1. $\frac{1}{80}$
  2. $\frac{7}{16}$
  3. $0$
  4. $1$
Answer
  1. 0
Solution:
Total number of bulbs = 80
Let E of the event that bulb selected at random form the lot has life time 1150hrs.
We see from the frequency table given above that none of the bulb has life time 1150hrs.
$\therefore\ \text{P(E)}=\frac{0}{80}=0$
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Question 251 Mark
Write the correct answer in the following:
There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is:
  1. 46.5
  2. 49.5
  3. 53.5
  4. 56.5
Answer
  1. 56.5
Solution:
Given, n = 50, then mean $\bar{\text{x}}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}$
Then, $\bar{\text{x}}=\frac{1}{50}\times\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}$
$\Rightarrow\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}=50\bar{\text{x}}$
Now, subtract each observation from 53, we get a new mean say $\bar{\text{x}}_\text{new}$
$\therefore\ \bar{\text{x}}_\text{new}=\frac{(-\text{x}_1+53)+(-\text{x}_2+53)\ +\ .....\ +(-\text{x}_{50}+53)}{50}$
$\Rightarrow-3.5=\frac{-(\text{x}_1+\text{x}_2\ +\ .....\ +\text{x}_{50})+(53+53\ +\ ....\ +50\text{times})}{50}$
$\Rightarrow-3.5\times50=(-\text{x}_1+\text{x}_2+\ ....\ +\text{x}_{50})+53\times50$
$\therefore$ Mean of 50 observation $=\frac{1}{50}\sum\limits^{50}_{\text{i}=1}\text{x}_{\text{i}}$
$=\frac{1}{50}\times2852=56.5$ $\Bigg[\because\text{ mean}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}\Bigg]$
Hence, the mean of given number is 56.5
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Question 261 Mark
Write the correct answer in the following:
If the mean of the observations: x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of the last three observations is:
  1. $10\frac{1}{3}$
  2. $10\frac{2}{3}$
  3. $11\frac{1}{3}$
  4. $11\frac{2}{3}$
Answer
  1. $11\frac{1}{3}$
Solution:
Given that, the mean of the observation x, x + 3, x + 5, x + 7 and x + 10 is 9.
$\therefore\ \frac{\text{x}+\text{x}+3+\text{x}+5+\text{x}+7+\text{x}+10}{5}=9$
$\Rightarrow {\text{5x}}+25=45$
$\Rightarrow\text{5x}=20$
$\Rightarrow \text{x}=4$
$\therefore$ Last three odservations are x + 5 = 4 + 5 = 9, x + 7 = 4 + 7 = 11
and x + 10 = 4 + 10 = 14
So, the mean of last three observations $=\frac{9+11+14}{3}=\frac{34}{3}=11\frac{1}{3}$
Hence, the mean of last three observation is $11\frac{1}{3}.$
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Question 271 Mark
Write the correct answer in the following:
The class mark of the class 90-120 is:
  1. 90
  2. 105
  3. 115
  4. 120
Answer
  1. 105
Solution:
$\text{Class mark }=\frac{\text{Upperlimit + Lowerlimit}}{2}$
$\Rightarrow \text{Class mark}=\frac{90+120}{2}=\frac{210}{2}=105$
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MCQ 281 Mark
Write the correct answer in the following: The mean of five numbers is $30.$ If one number is excluded, their mean becomes $28.$ The excluded number is:
  • A
    $28$
  • B
    $30$
  • C
    $35$
  • ✓
    $38$
Answer
Correct option: D.
$38$
Let $x_1,_{ }x_2,_{ }x_3, x_4,$ and $x_{5 }$ be five numbers and one of the excluded number be $x_5,$
Given, mean of the numbers $= 30$
$\Rightarrow\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}{5}=30$
$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}=150$
$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}=150-\text{x}_5$
On dividing both sides by $4$, we get
$\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}{4}=\frac{150-\text{x}_5}{4}\ \dots(\text{i})$
Given, mean of the numbers = 28
$\therefore\ \frac{150-\text{x}_5}{4}=28 [$from $Eq. (i)]$
$\Rightarrow 150 -\text{x}_5=112$
$\Rightarrow \text{x}_5=150 -112$
$\Rightarrow\text{x}_5=38$
Hence, the excluded number is $38.$
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Question 291 Mark
Write the correct answer in the following:
Two coins are tossed 1000 times and the outcomes are recorded as below:
Number of heads
2
1
0
Frequency
200
550
250
Based on this information, the probability for at most one head is:
  1. $\frac{1}{5}$
  2. $\frac{1}{4}$
  3. $\frac{4}{5}$
  4. $\frac{3}{4}$
Answer
  1. $\frac{4}{5}$
Solution:
The total number of coins tossed, n(S) = 1000
Number of outcomes in which atmost one head, n(E) = 550 + 250 = 800
$=\frac{\text{n(E)}}{\text{n(S)}}=\frac{800}{1000}=\frac{4}{5}$
Hence, the probability for atmost one head is $\frac{4}{5}$
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MCQ 301 Mark
If $\bar{\text{x}}$ is the mean of $x_1, x_2, ......, x_n,$ then for $\text{a}\ne0,$ then mean of $\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n},\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}$ is:
  • A
    $\Big(\text{a}+\frac{1}{\text{a}}\Big)\bar{\text{x}}$
  • B
    $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{2}$
  • C
    $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{\text{n}}$
  • ✓
    $\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\bar{\text{x}}}{2\text{n}}$
Answer
Correct option: D.
$\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\bar{\text{x}}}{2\text{n}}$
Given, mean of $x_1, x_2, ......, x_n$ is $\bar{\text{x}}$
$\therefore\ \sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}=\text{n}\bar{\text{x}}$
Now, let the mean of $\Big(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n},\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)$ is $\bar{\text{z}}$
Then, $\bar{\text{z}}=\frac{(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n})+\Big(\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)}{\text{n}+\text{n}}$
$=\frac{\text{a}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})+\frac{1}{\text{a}}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\cdot\text{n}\bar{\text{x}}}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big){\bar{\text{x}}}}{2}$
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