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(C) 0.75

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.

∴ Number of children who do not like to eat potato chips, n(E) = 364 – 91 = 273

∴ Probability that he/she does not like to eat potato chips  = n(E)/n(S) = 273/364 = 0.75

Hence, the probability that he/she does not like to eat potato chips is 0.75.

(a) frequency polygon

Arranging the data in ascending order 05, 21, 36, 42, 56, 75, 81

Number of terms in the data = 7 – odd

∴ (\(\frac{n+1}{2})^{th}\) term = \(\frac {7+1}{2}\) = 8/2 = 4th term is the median = 42.

Given data in ascending order: 

35, 35, 37, 37, 37, 37, 40, 42, 42, 43 

∴ The observation repeated maximum number of times = 37 

∴ Mode of the given data is 37 kg

No it is not correct that in a family having three children, there may be no girl, one girl, two girls or three girls, the probability of each is ¼. .

Let boys be B and girls be G

Outcomes can be BBB , GGG , BBG , BGB , GBB, GGB, GBG , BGG

Then Probability of 3 girls = 1/8

Probability of 0 girls = 1/8

Probability of 2 girls = 3/8

Probability of 1 girl = 3/8

(D) The answer is 30

Let the upper class limit be x and the lower class limit be y. 

Class mark = 10 …[Given] 

Class-mark = (Lower class limit + Upper class limit)/2

∴ 10 = (x + y)/2

∴ x + y = 20 …(i) 

Class width = 6 … [Given] 

Class width = Upper class limit – Lower class limit 

∴ x – y = 6 …(ii) 

Adding equations (i) and (ii), 

x + y = 20 

x – y = 6

2x = 26 

∴ x = 13 

Substituting x = 13 in equation (i), 

13 + y = 20 

∴ y = 20 – 13 

∴ y = 7 

∴ The required class is 7 – 13.

Total appropriate answers received = 19 + 36 + 10 = 65 

Majority of people’s opinion is prohibition in public places only i.e., B.

Σf_i = 84

Σf_ix_i = 144

\(\overline x = \frac {\sum f_ix_i}{\sum f_i} = \frac {144}{84}\)

Mean = 1.714285

The least value of observations = 170 

The height value of observations = 724

 Σf_i = 200 

Σf_i x_i = 17000

\(\overline x = \frac {\sum f_ix_i} {\sum f_i} = \frac {17000}{200} = \frac {170}{2}\)

\(\overline x \) = 85

Mean = 85

Arranging the data in the from of a frequency table, we have :

Since the value 120 occurs maximum number of times i.e 4. Hence, the modal value is 120.

We have, 

N = 100

\(\frac{N}{2}=\frac{100}{2}=50\)

The cumulative frequency just greater than \(\frac{N}{2}\) s 67 then the median class 94.5-104.5 such that,

l = 94.5, f = 33, F = 34, h = 104.5 – 94.5 = 10

Median = I + \(\frac{\frac{N}{2}-F}{f}\times h\)

\(=94.5+\frac{50-34}{33}\times10\)

\(=94.5+\frac{16}{33}\times10\)

\(=94.5+4.85=99.35\)

From the data its observed that,

Model shirt size = 40 since it was the size which occurred for the maximum number of times.

Given data in ascending order: 

60, 90, 95, 99, 100, 100, 100 

Here, the observation repeated maximum number of times = 100 

∴ The mode of the given data is 100.

Three fair coins are tossed together

Sample spade = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

n(S) = 8

(i) Let A be the event of getting all heads

A = {HHH}

n(A) = 1

P(A) = \(\frac{n(A)}{n(S)}=\frac{1}{8}\)

(ii) Let B be the event of getting atleast one tail.

B = {HHT, HTH, HTT, THH, THT, TTH, TTT}

n(B) = 7

P(B) = \(\frac{n(B)}{n(S)}=\frac{7}{8}\)

(iii) Let C be the event of getting atmost one head

C = {HTT, THT, TTH, TTT}

n(C) = 4

P(C) = \(\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}\)

(iv) Let D be the event of getting atmost two tails.

D = {HTT, TTT, TTH, THT, THH, HHT, HTH}

n(D) = 7

P(D) = \(\frac{n(D)}{n(S)}=\frac{7}{8}\)

(2) \(\frac{p}{p\,+\,q\,+\,r}\)

Correct option is: A) Mode = 3 Median – 2 Mean

Correct option is: C) \(\frac{\overline{X}}a{}\)

First n natural numbers are

1, 2, 3, 4, …, n

We know that mean or average of observations, is the sum of the values of all the observations divided by the total number of observations.

and, we have given series

1, 2, 3, …, n

Clearly the above series is an AP(Arithmetic progression) with

first term, a = 1 and

common difference, d = 1

And no of terms is clearly n.

And last term is also n.

We know, sum of terms of an AP if first and last terms are known is:

\(S_n=\frac{n}{2}(a+a_n)\)

Putting the values in above equation we have sum of series i.e.

\(1+2+3+...+n=\frac{n}{2}(1+n)\)

\(=\frac{n(n+1)}{2}...[1]\)

As,

Mean \(=\frac{sum\,of\,all\,terms}{no\,of\,terms}\)

\(=\frac{1+2+3+...+n}{n}\)

⇒ Mean \(=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}\)

Given, mean = 15

\(\Rightarrow \frac{n+1}{2}=15\)

⇒ n + 1 = 30 

⇒ n = 29

Correct option is: A) 0

Mean = \(\frac {sum \, of\, obs}{Total \, No.\, of\, obs}\) 

= \(\frac {-3+ (-2) + (-1) + 0 + 1+2 +3}{7}\) = \(\frac 07\) = 0

Correct option is: A) 0

Correct option is: B) 19

Given that the mean of data is 10.

Let number of observations be n

\(\therefore\) Sum of original observations is \(\sum x_i = 10n\) ...(1)

Given that each observation (\(x_i\)) is multiplied by 2 and 1 is subtracted from each result.

\(\therefore\) New observations are of type (2\(x_i\)-1).

Sum of new observations = \(\sum\) (2\(x_i\)-1).

= 2\(\sum\)\(x_i\) -\(\sum\) 1

= 20n - n (\(\because\) \(\sum\)\(x_i\) = 10n & \(\sum\) 1 = n)

= 19 n

\(\therefore\) Mean of new observations = \(\frac {sum \, of \, new\,observations}{ number \, of\, observations}\)

= \(\frac {19n}{n} = 19\) 

Hence, the mean of new observations is 19.

Correct option is: B) 19

Correct option is: B) 15.5

Correct option is: C) 28

Let \(x_1\)'s are observations of the original data .

Given that mean of data is 9.

Let total no. of observation be n.

\(\therefore\) Sum of observations = n\(\overline x\) = 9n

\(\Rightarrow\) \(\sum \limits _{i=1}^n x_i\) = 9n ....(1)

Since, each observation ( \(x_1\)'s) are multiplied by 3 and then 1 is added to each result.

\(\therefore\) (\(3x_i + 1\))'s are new observations.

\(\therefore\) Mean of formed data = \(\frac {sum\, of \,new \,observations}n\)

= \(\frac {\sum \limits_{i=1}^n (3x_i+1)}{n}\)

= \(\frac {\sum \limits_{i=1}^n 3x_i + \sum \limits_{i=1}^n 1}{n}\)

= \(\frac {3\sum \limits_{i=1}^n x_i +n}{n}\) (\(\because\) \(\sum \limits_{i=1}^n 1 = 1 + 1+1+...+1 (n \, times) = n\))

= \(\frac {3\times 9n+n}{n} \) (From (i))

= \(\frac {28n}{n} = 28\) 

Hence, the mean of new observations is 28.

Correct option is: C) 28

Correct option is (D) 27.4

New mean \(=2\times12.2+3\)

= 24.4 + 3 = 27.4

Correct option is  D) 27.4

Correct option is (C) 12.6

If 5 is added to each observation then new mean is increased by 5 than the previous mean of the data.

\(\therefore\) New mean = 7.6+5

= 12.6

Correct option is  C) 12.6

Correct option is (B) 45

Given that mean of the data \(x_1, x_2, x_3 , x_4 , x_5\) is 15.

\(\therefore\) \(\frac{x_1+x_2+x_3+x_4+x_5}5=15\)

\(\Rightarrow\) \(x_1+x_2+x_3+x_4+x_5\) \(=15\times5=75\)   _________(1)

If each observation is multiplied by 3 then

Mean \(=\frac{3x_1+3x_2+3x_3+3x_4+3x_5}5\)

\(=\frac35(x_1+x_2+x_3+x_4+x_5)\)

\(=\frac35\times75=45\)

Correct option is B) 45

\(Mean = \cfrac{The\,sum \,of\, all \,observatiojns }{Total \,number \,of \,observations}\) 

∴ The sum of all observations = Mean x Total number of observations 

The mean of 5 observations is 50 

Sum of 5 observations = 50 x 5 = 250 …(i)

One observation was removed and mean of remaining data is 45.

Total number of observations after removing one observation = 5 – 1 = 4 

Now, mean of 4 observations is 45. 

∴ Sum of 4 observations = 45 x 4 = 180 …(ii) 

∴ Observation which was removed = Sum of 5 observations – Sum of 4 observations = 250 – 180 … [From (i) and (ii)] 

= 70 

∴ The observation which was removed is 70.

Given data in ascending order : 

2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20. 

∴ Number if observations (n) = 10 (i.e., even) 

∴ Median is the average of middle two observations Here, the 5th and 6th numbers are in the middle position.

∴ Median = \(\frac{(x+1)\,+\,(x+3)}{2}\)

∴ 11 = \(\frac{2x \,+ \,4}{2}\)

∴ 22 = 2x + 4 

∴ 22 – 4 = 2x 

∴ 18 = 2x 

∴ x = 9

\(Mean = \cfrac{The\,sum \,of\, all \,observatiojns }{Total \,number \,of \,observations}\)

∴ The sum of all observations

 = Mean x Total number of observations 

The mean of 35 observations is 20 

∴ Sum of 35 observations = 20 x 35 = 700 ...(i) 

The mean of first 18 observations is 15 

Sum of first 18 observations =15 x 18 = 270 …(ii) 

The mean of last 18 observations is 25 Sum of last 18 observations = 25 x 18

= 450 …(iii) 

∴ 18^th observation = (Sum of first 18 observations + Sum of last 18 observations) – (Sum of 35 observations)

= (270 + 450) – (700) … [From (i), (ii) and (iii)] 

= 720 – 700 = 20 

The 18^th observation is 20.

Given that A is twice as likely to win as B. 

Therefore, A : B = 2 : 1

(or) A : B = 4 : 2 ... (1) 

Also given that B is twice as likely to win as C. 

Therefore, B : C = 2 : 1 … (2) 

From (1) and (2), 

A : B : C = 4 : 2 : 1 

∴ A = 4k, B = 2k, C = 1, where C is a constant of proportionality. 

Probability of A wining is \(\frac{4k}{7k}=\frac{4}{7}\)

Probability of B wining is \(\frac{2k}{7k}=\frac{2}{7}\)

Probability of C wining is \(\frac{k}{7k}=\frac{1}{7}\)

Let S be the sample when a die is thrown.

Then S = {1, 2, 3, 4, 5, 6}, n(S) = 6 

Let A be the event of getting a prime number. 

A = {2, 3, 5}, n(A) = 3 

Let B be the event of getting a number greater than or equal to 3. 

B = {3, 4, 5, 6}, n(B) = 4

(i) P(a prime number) = \(\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

(ii) P(a number ≥ 3) = \(\frac{n(B)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

Some of the examples of data that we can collect from our day-to-day life are as follows : 

(i) Number of TV viewers in the city. 

(ii) Number of Colleges in the city. 

(iii) Number of sugar factories in the city. 

(iv) Measuring height of students in the class-room. 

(v) Number of children below 15 years in India.

Answer is (2) 3p