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11 months ago

Statistics Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let the mean and the variance of 20 observations x1, x2,…….x20 be 15 and 9, respectively. For $\in R$ , if the mean of ${(x_{1} + α)}^{2}, {(x_{2} + α)}^{2}, . . . ., {(x_{20} + α)}^{2}$ is 178, then the square of the maximum value of is equal to………….

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Vishal Baghel

Contributor-Level 10

$\bar{x} = 15 \Rightarrow \sum_{i = 1}^{20} x_{i} = 300$

$\frac{\sum_{i = 1}^{20} {(x_{i} - 15)}^{2}}{20} = 9 \Rightarrow \sum_{i = 1}^{20} {(x_{i} - 15)}^{2} = 180$

$\sum_{i = 1}^{20} {(x_{i} + α)}^{2} = 178 * 20 = 3560$

$4680 + 2 α (300) + 20 α^{2} = 3560$

$α^{2} + 30 α + 234 - 178 = 0$

= 2, 28

$α_{m a x} 2 = {(- 2)}^{2} = 4$

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New answer posted

11 months ago

Statistics Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

If the variance of the following frequency distribution :

$\begin{array}{l} Class & : & 10 - 20 & 20 - 30 & 30 - 40 \\ Frequency & : & 2 & x & 2 \end{array}$ is 50 , then $x$ is equal to:

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alok kumar singh

Contributor-Level 10

Since, ${l i m}_{x \to 0} ? \frac{f (x)}{x}$ exist $\Rightarrow f (0) = 0$

Now, $f^{'} (x) = {l i m}_{h \to 0} ? \frac{f (x + h) - f (x)}{h} = {l i m}_{h \to 0} ? \frac{f (h) + x h^{2} + x^{2} h}{h} ($ take $y = h)$

$= {l i m}_{h \to 0} ? \frac{f (h)}{h} + {l i m}_{h \to . 0} ? (x h) + x^{2}$

$\Rightarrow f^{'} (x) = 1 + 0 + x^{2}$

$\Rightarrow f^{'} (3) = 10$

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New answer posted

12 months ago

Statistics Maths NCERT Exemplar Solutions Class 12th Chapter Eight

The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and $\frac{9}{4}$ respectively. Then $x^{4} + y^{2}$ is equal to.

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Payal Gupta

Contributor-Level 10

$= \frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$

$∴ x + y = 12$ …. (i)

And variance

$= \frac{2^{2} + 1^{2} + 0^{2} + 0^{2} + 1^{2} + 2^{2} + {(x - 6)}^{2} + {(y - 6)}^{2}}{8}$

$= \frac{9}{4}$

$∴ {(x - 6)}^{2} + {(y - 6)}^{2} = 8$ ……. (ii)

From (i) and (ii)

x = 4 and y = 8

$∴ x^{4} + y^{2} = 320$

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New answer posted

a year ago

Statistics Class 11th

34. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

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Payal Gupta

Contributor-Level 10

34. Given, n = 100.

incorrect mean ( $\bar{x}$ ) = 20.

incorrect standard deviation (σ) = 3

We know that,

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} n_{i}$

$\Rightarrow 20 = \frac{1}{100} \sum_{i = 1}^{n} n_{i}$

$\Rightarrow \sum_{i = 1}^{n} n_{i} = 2000 .$

So, incorrect sum of observation = 2000

Correct sum of observation $= 2000 - 21 - 21 - 18$

= 1940

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New answer posted

a year ago

Statistics Class 11th

33. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

Subject Mathematics Physics Chemistry

Mean

...more

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Payal Gupta

Contributor-Level 10

33. C.V in mathematics = $\frac{12}{42} * 100 = 28.57 .$

C.V in Physics = $\frac{15}{32} * 100 = 46.87 .$

C.V in chemistry = $\frac{20}{40.9} * 100 = 48.89$

$∴$ Chemistry has the highest variability and mathematics has the lowest variability.

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New answer posted

a year ago

Statistics Class 11th

32. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted.

(ii) If it is replaced by 12.

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Payal Gupta

Contributor-Level 10

32. (i) Given, n = 20.

Incorrect mean $(\bar{x}) = 10$

Incorrect standard deviation $(σ) = 2$

We know that,

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$

$\Rightarrow 10 = \frac{1}{20} \sum_{i = 1}^{20} x_{i}$

$\Rightarrow \sum_{i = 1}^{20} x_{i} = 200$

So, incorrect sum of observation = 200.

correct sum of observation =200 – 8 = 192

And correct mean $= \frac{192}{19} = 10.1$

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New answer posted

a year ago

Statistics Class 11th

31. Given that $\bar{x}$ is the mean and σ2 is the variance of n observations x₁, x₂, .,xn. Prove that the mean and variance of the observations ax₁, ax₂, ax₃, ., ax_n are a $\bar{x}$ and a₂ σ₂, respectively, (a ≠ 0).

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Payal Gupta

Contributor-Level 10

31. For n observations x₁, x₂,……., x_n .

We have mean = $\bar{x} = \frac{\sum_{i = 1}^{n} x_{i}}{n} (1)$

and variance = $σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} (2)$

Let y_i be the new observations with same n.

So, y_i = ax_i (3)

Now mean, $\bar{y} = \frac{\sum_{i = 1}^{n} y_{i}}{n} = \frac{\sum_{i = 1}^{n} a x_{i}}{n} = \frac{a \sum_{i = 1}^{n} y_{i}}{n} = a \bar{x} [(1)]$

So $\bar{y} = a \bar{x} (4)$

And, putting (3) and (4) in (2) we get,

$σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(\frac{y_{i}}{a} - \frac{\bar{y}}{a})}^{2}$

$\Rightarrow σ^{2} = \frac{1}{a^{2}} [\frac{1}{n} \sum_{i = 1}^{n} {(y_{i} - \bar{y})}^{2}]$

$\Rightarrow {(σ^{'})}^{2} = σ^{2} α^{2} .$

Hence, the mean and variance of ax₁, ax₂, ……, ax_n are $a \bar{x}$ and a² σ² .

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New answer posted

a year ago

Statistics Class 11th

30. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

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Payal Gupta

Contributor-Level 10

30. Let 'x' be the given observations with n = 6.

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New answer posted

a year ago

Statistics Class 11th

27. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

$\sum_{i = 1}^{50} = x_{i} = 212, \sum_{i = 1}^{50} x_{i}^{2} = 902.8, \sum_{i = 1}^{50} y_{i} = 261, \sum_{i = 1}^{50} {y_{i}}^{2} = 1457.6$

Which is more varying, the length or weight?

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

27. Given, n=50.

$\begin{array}{l} \sum_{i = 1}^{n} x i = 212 \sum_{i = 1}^{50} x i^{2} = 902.8 \end{array}$

$\sum_{i = 1}^{50} y i = 261 \sum_{i = 1}^{50} y i^{2} = 1457.6$

So, $\bar{x} = \frac{\sum_{i = 1}^{50} x_{i}}{n} = \frac{212}{50} = 4.24$

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New answer posted

a year ago

Statistics Class 11th

26. The following is the record of goals scored by team A in a football session:

No. of goals scored	0	1	2	3	4
No. of matches	1	9	7	5	3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

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Payal Gupta

Contributor-Level 10

26. For team A.

Hence, we conclude that team A is more consistent.

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