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C) Harsha and Ritu are correct.

Ascending order: 4, 4, 7, 9, 10, 13, 25, 31, 34, 40, 42.

Descending order : 42, 40, 34, 31, 25, 13, 10, 9, 7, 4, 4

(i) Monthly wages of firm A = Rs 5253

Number of wage earners in firm A = 586

∴Total amount paid = Rs 5253 × 586

Monthly wages of firm B = Rs 5253

Number of wage earners in firm B = 648

∴Total amount paid = Rs 5253 × 648

Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A (σ₁² ) = 100

∴ Standard deviation of the distribution of wages in firm

A (( σ₁ ) =  √100 = 10

Variance of the distribution of wages in firm = B ((σ₂²  )) = 121

∴ Standard deviation of the distribution of wages in firm B (σ₂² ) = √121 = 11

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.

Thus, firm B has greater variability in the individual wages.

Here SD(Boys) is 3 and SD (girls) = 2 

Coefficient variability = \(\frac{SD}{Mean}\times100\) 

Coefficient variance (Boys) = \(\frac{3}{60}\times 100\) 

Cv (Boys) = 5 

Coefficient variance (Girls) = \(\frac{2}{45}\times 100\) 

Cv (Boys) = 5 

Coefficient variance (Girls) =  \(\frac{2}{45}\times 100\) 

Cv (Girls) = 4.4

Hence, Cv Boys is greater than Cv girls,then the distribution of weights of boys is more variable than that of girls

In case of heights,

Mean = 63.2 inches and SD = 11.5 inches. 

So, the coefficient of variation,

CV = \(\frac{SD}{Mean}\times100\) 

CV =   \(\frac{11.5}{63.2}\times100\)  = 18.196

In case of weights,

Mean = 63.2 inches and SD = 5.6 inches.

So, the coefficient of variation,

CV = \(\frac{SD}{Mean}\times100\) 

CV =   \(\frac{5.6}{63.2}\times100\)  = 8.86

CV of heights > CV of weights 

So, heights show more variability

(i) Average weekly wages = \(\frac{Total\,weekly\,wages} {No.\,of \,workers}\)

Total weekly wages = (Avg weekly wages) x (No. of workers) 

Total weekly wages of Firm A = 52.5 x 586 = Rs 30765 

Total weekly wages of Firm B = 47.5 x 648 = Rs 30780

Firm B pays a larger amount as Firm A

(ii) Here SD(firm A) 10 and SD (Firm B) = 11 

Coefficient variance (Firm A) = \(\frac{10}{52.5}\times 100\) 

Cv (Firm A) = 19.04

Coefficient variance (Firm B) = \(\frac{11}{47.5}\times100\) 

Cv (Firm B) = 23.15 

Hence, Cv of firm B is greater that that of firm A, Firm B has greater variability in individual wages.

(i) Both the factories pay the same mean monthly wages. 

For factory A there are 560 workers. And for factory B there are 650 workers. 

So, factory A totally pays as monthly wage = (5460 x 560) Rs. 

= 3057600 Rs. 

Factory B totally pays as monthly wage = (5460 x 650) Rs. 

= 3549000 Rs. 

That means, factory B pays a larger amount as monthly wages. 

(ii) Mean wages of both the factories are the same, i.e., Rs. 5460. 

To compare variation, we need to find out the coefficient of variation (CV).

We know, CV = \(\frac{SD}{Mean}\times100\), where SD is the standard deviation. 

The variance of factory A is 100 and the variance of factory B is 121. 

Now, SD of factory A = 

\(\sqrt{100}\) = 10

And, SD of factory B =

 \(\sqrt{121}\) = 11

Therefore, 

The CV of factory A = 

\(\frac{10}{5460}\times100\) = .183

The CV of factory B =

\(\frac{11}{5460}\times100\) = .201

Here, the CV of factory B is greater than the CV of factory A. 

Hence, factory B shows greater variability.

Variation of the distribution of wages in plant A (σ² =18)

So, Standard deviation of the distribution A (σ – 9)

Similarly, the Variation of the distribution of wages in plant B (σ² =100)

So, Standard deviation of the distribution B (σ – 10)

And, Average monthly wages in both the plants is 2500,

Since, the plant with a greater value of SD will have more variability in salary.

∴ Plant B has more variability in individual wages than plant A

Mean wages of both the factories are the same, i.e., Rs. 5300. 

To compare variation, we need to find out the coefficient of variation (CV)

We know, CV = , \(\frac{SD}{Mean}\times100\) where SD is the standard deviation. 

The variance of factory A is 100 and the variance of factory B is 81. 

Now, SD of factory A = \(\sqrt{100}\) = 10

And, SD of factory B =   \(\sqrt{81}\) = 9

Therefore, 

The CV of factory A =   \(\frac{10}{5300}\times100\) = 0.189

The CV of factory B = \(\frac{9}{5300}\times100\) = 0.169

Here, the CV of factory A is greater than the CV of factory B. 

Hence, factory A has more variation.

Mean wages of both the factories = Rs. 5300

Find the coefficient of variation (CV) to compare the variation.

We know, CV = SD/Mean x 100, where SD is the standard deviation.

The variance of factory A is 100 and the variance of factory B is 81.

Now, SD of factory A = √100 = 10

And, SD of factory B = √81 = 9

Therefore,

The CV of factory A = 10/5300 x 100 = 0.189

The CV of factory B = 9/5300 x 100 = 0.169

Here, the CV of factory A is greater than the CV of factory B.

Hence, factory A has more variation in wages.

Variation of the distribution of wages in plant A(σ² =18) 

So, Standard deviation of the distribution A(σ - 9) 

Similarly, the Variation of the distribution of wages in plant B(σ² =100) 

So, Standard deviation of the distribution B(σ - 10) 

And, Average monthly wages in both the plants is 2500, 

Since The plant with a greater value of SD will have more variability in salary. 

Hence, Plant B has more variability in individual wages than plant A

A and B are two mutually exclusive events of a random experiment.

P(not A) = 0.45,

P(A) = 1 – P(not A)

P(A∪B) = 0.65 = 1 – 0.45 = 0.55

P(A∪B) = P(A) + P(B) = 0.65

0.55 + P(B) = 0.65

P(B) = 0.65 – 0.55 

= 0.10

Here P (A ∪ B) = 0.6, P (A ∩ B) = 0.2

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

0.6 = P (A) + P (B) – 0.2

P(A) + P(B) = 0.8

P(\(\bar{A}\)) + P(\(\bar{B}\)) = 1 – P(A) + 1 – P(B)

= 2 – [P(A) + P(B)]

= 2 – 0.8

= 1.2

P(A) = 0.5 

Since A and B are mutually inclusive events

P(B) = 0.3 events.

P(\(\bar{A}\))∪P(\(\bar{B}\)) = 1 – [P(A) + P(B)]

= 1 – [0.5 + 0.3] = 0.2

(b) P(A ∩ B) = P(A) x P(B)

(c) \(\frac{P(A∩B)}{P(A)}\)

(d) Independent

(d) \(\frac{1}{4}\)

Number of spade cards is 13. 

Total number of cards in pack = 52 

Probability of drawing a spade card is = \(\frac{13}{52}=\frac{1}{4}\)

(a) 10

The ascending order of 10, 14, 11, 9, 8, 12, 6 is 6, 8, 9, 10, 11, 12, 14.

In this order middle number is 10. 

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

= \((\frac{7+1}{2})^{th}\)

= 10 

∴ Median 10.

(d) 13 

The values are in ascending, order. 

∴ The mean is the middle value.

(a) Speed or rates

(b) lower quartile

Given,

Mean = 20.6

\(\frac{\sum yx}{N}=20.6\)

\(\frac{530+25p}{50}=20.6\)

530 + 25p = 20.6 (50)

25p = 20.6 (50) – 530

p = \(\frac{500}{25}=20\) 

Let the number of candidates from school III be P.

Given,

Average score of all schools = 66

\(\frac{N1x1+N2x2+N3x3+N4x4}{N1+N2+N3+N4}=66\)

\(\frac{60\times75+48\times80+p\times55+40\times50}{+0+48+p+40}=66\)

\(\frac{4500+3840+55p+2000}{148+p}=66\)

10340 + 55P = 66P + 9768

10340 – 9768 = 66P – 55P

p = \(\frac{572}{11}=52\)

Correct option is (B) 6

Observations in ascending order are 3, 4, 6, 7 and 12.

Number of observations = 5 (odd)

\(\therefore\) Median \(=(\frac{5+1}2)^{th}\) observation

\(=3^{rd}\) observation = 6

Correct option is  B) 6

Correct option is (C) 14.4

\(\bar x=A+\frac{\sum f_id_i}{\sum f_i}\)

\(=15-\frac{12}{20}\)

= 15 - 0.6

= 14.4

Correct option is  C) 14.4

Terms are 6, 7, x - 2, x, 17, 20 

No of terms = 6

We know that, if even no of terms or observations are given, then the median of data is mean of the values of \((\frac{n}{2})^{th}\) term and \((\frac{n}{2}+1)^{th}\) term. Where n is no of terms.

In this case, n = 6

\(\frac{n}{2}=3\) and \(\frac{n}{2}+1=4\)

i.e. median of above data is mean of 3^rd and 4^th term

⇒ median \(=\frac{(\text{x-2})+\text{x}}{2}=\frac{\text{2x-2}}{2}\)

⇒ 16 = x - 1 [As median is 16]

⇒ x = 17

We know that empirical relation between mean, median and mode is 

Mode = 3 Median - 2 Mean

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 \(=\frac{5n}{9}\)

\(\Rightarrow \frac{n+1}{2}=\frac{5n}{9}\)

⇒ 9n + 9 = 10n 

⇒ n = 9