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Answer is (D) 7.5

Answer is (B) by histogram

Answer is (C) 3

\(\mu\) = mean = 2000

\(\sigma\) = Standard deviation = 150

\(X \sim N(\mu, \sigma^2)\)

\(\frac{X - \mu}{\sigma} \sim N(0, 1)\)

⇒ \(\frac{X - 2000}{150} \sim N(0, 1)\)

\(\because 1750 \le X \le 2250\)

⇒ \(\frac {1750 - 2000}{150} \le \frac{X - 2000}{150} \le \frac{2210-2000}{150}\)

⇒ \(\frac{-250}{150} \le \frac{X - 2000}{150} \le \frac{250}{150}\)

⇒ \(\frac{-5}3 \le \frac {X - 2000}{150} \le \frac53\)

\(\because \frac{X - 2000}{150} \sim N(0, 1)\)

\(\therefore P\left(\frac{-5}3 \le \frac{X - 2000}{150} \le \frac 53\right) = P(Y \le \frac53) - P(Y \le \frac{-5}3 )\)

\(= 0.9516 - 0.0485\)

\(= 0.9031\)

The number of workers whose wages lie between 1750 & 2250 = 0.9031 x total no. of workers 

\(= 0.9031 \times 1000\)

\(= 903.1 \approx 903\)  (approx)

Correct option is: B) 12

Given that median of data 6, 10, 20, x, 12, 14 is 12.

(A) If x = 6,

then ascending order of observation is as follows :

6, 6, 1 0 > 12, 14, 20 total 6 observations

\(\therefore\) Median = \(\frac {\frac n2th(\frac n2 + 1)^{th}}{2} = \frac {3^{rd} + 4^{th}}{2} = \frac {10+12}{2} = \frac {22}2 = 11\)

(B) If x = 12

Then ascending order of observations is as follows :

6, 10, 12, 12, 14, 20

\(\therefore\) Median = \(\frac {12 + 12}{2} = \frac {24}2 =12\)

(C) If x = 4

Then ascending order of observation is as follows:

4, 6, 10, 12, 14, 20.

\(\therefore\) Median = \(\frac {10 + 12}{2} = \frac {22}2 =11\)

(D) If x = 14

Then ascending order of observation is as follows:

 6, 10, 12, 14, 14, 20.

\(\therefore\) Median = \(\frac {12 + 14}{2} = \frac {26}2 =13\)

Correct option is: B) 12

Correct option is: C) 5.6

First 5 prime numbers are 2, 3, 5, 7 and 11.

Sum of first 5 prime numbers = 2 + 3 + 5 + 7 + 11 = 28

\(\therefore\) Mean = \(\frac {Sum}{Total number} = \frac {28}5 = 5.6\)

Correct option is: C) 5.6

Given, 

N = 2a,

Mean,

\(\bar{x}=\frac{a+a+...+a+(-a)+(-a)+...(-a)}{2n}\)

\(=\frac{0}{2n}\)

 = 0

Standard deviation, 
\(σ=2\)

⇒ \(σ^2=2^2\)

\(⇒\frac{\displaystyle\sum {x_i}^2}{N}-(\frac{\displaystyle\sum {x_i}}{N})^2=4\)

\(⇒\frac{\displaystyle\sum {x_i}^2}{N}-x^{-2}=4\)

\(⇒\frac{\displaystyle\sum {x_i}^2}{N}-0=4\) 

\(⇒\displaystyle\sum {x_i}^2=8n\)

⇒ a² + a² + ⋯ + a² (to 2n terms) 2a² = 8n

⇒ a² = 4

⇒ a = ± 2

∴ |a| = |± 2|

= 2

B. less than 0

We know that the probability expressed as a percentage always lie between 0 and 100. So, it cannot be less than 0.

No, since the number of coin increases, the ratio of the number of heads to the total number of tosses will be nearer to ½ but not exactly ½.

According to the question,

Mean = 20.2

So,

∑f_ix_i/∑f_i = 20.2

So we get,

∑f_ix_i = (10×6)+(15×8)+(20×p)+(25×10+)(30×6)

=610+20p

We also get,

∑f_i=6+8+p+10+6

=30+p

Therefore,

∑f_ix_i/∑f_i=20.2

(610+20p)/30+p=20.2

20.2(30+p) =610+20p

606+20.2p=610+20p

20.2p-20p=610-606

0.2p=4

p=4/0.2

p=20

Given,

Mean = 16.6

\(\frac{\sum yx}{N}=16.6\)

\(\frac{24p+1228}{100}=16.6\)

24p + 1228 = 1660

24p = 1660 – 1228

24p = 432

p = \(\frac{432}{24}=18\)

We may calculate class marks (x_i) for each interval by using the relation

x_i \(=\frac{Upper\,class\,limit+lower\,class\,limit}{2}\)

Class size = 0.04 

Now taking 0.14 as assumed mean (A), we can calculate as follows:

Mean (x̅) = A + \(\frac{\sum f_iu_i}{N}\times h\)

\(=0.14+\frac{-31}{30}\times(0.04)\)

\(=0.14-0.04133\)

\(=0.099\)ppm

SO, 

mean concentration of SO₂ in the air is 0.099 ppm

Statistics is the science which deals with the collection, presentation, analysis and interpretation of numerical data.

(i) Numerical facts alone constitute data.

(ii) Qualitative characteristics like intelligence, poverty, etc., which cannot be measured numerically, do not form data.

(iii) Data are aggregate of facts. A single observation does not form data.

(iv) Data collected for a definite purpose may not be suited for another purpose.

(v) Data in different experiments are comparable.

Primary data: The data collected by the investigator himself with a definite plan in mind are known as primary data. These data are, therefore, highly reliable and relevant.

Secondary data: The data collected by someone, other than the investigator, are known as secondary data. Secondary data should be carefully used, since they are collected with a purpose different from that of the investigator and may not be fully relevant to the investigation.

(i) Any character which is capable of taking several different values is called a variate.

(ii) Each group into which the raw data is condensed, is called a class interval.

(iii) The difference between the true upper limit and the true lower limit of a class is called its class size.

(iv) The average of upper limit and lower limit of class interval is called as a class mark.

(v) Each class is bounded by two figures, which are called class limits.

(vi) In the exclusive form, the upper and lower limits of a class are respectively known as the true upper limit and true lower limit.

(vii) The number of times an observation occurs in a class is called the frequency.

(viii) The cumulative frequency corresponding to a class is the sum of all frequencies up to and including that class.

When an investigator collects data himself with a definite plan or design in his/her mind is called primary data. 

Data which are not originally collected rather obtained from published or unpublished source are known as secondary data.

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

= \(\frac{10\,+\,7\,+\,5\,+\,3\,+\,9\,+\,6\,+\,9}{7}\)

= \(\frac{49}{7}\)

Mean = 7  

The mean of yield per acre is 7 quintals.

Given data in ascending order: 

59, 68, 70, 74, 75, 75, 80 

∴ Number of observations(n) = 7 (i.e., odd) 

∴ Median is the middle most observation

Here, 4th number is at the middle position, which is = 74 

∴ The median of the given data is 74.

Arranging the data in ascending order: 

1, 2, 2, 2, 3, 3, 4, 5, 5, 6, 7, 7 

Here 2 occurs maximum number of times. 

∴ Mode is 2.

Arranging the given data in ascending order: 0, 1, 3, 3, 4, 7, 7, 16, 19 

Number of terms n = 9, which is odd. 

∴ Median = (\(\frac{n+1}{2}\))^th term

= (\(\frac{9+1}{2}\))^th term 

= (\(\frac{10}{2}\))^th term 

= 5^th term 

Hence Median = 4

1. 5.5 

2. 3,525 

3. 6 

4. 0

1. 5.5

2. 3525

3. 6

4. 0

(iii) 14

\(\frac{12+x+28}{3}\) = 18

x + 40 = 54 

x = 14