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Q 1.

Correct4

Incorrect-1

If the mean of the distribution is 2.6, then the value of y is Variatex|1|2|3|4|5| Frequency f of x|4|5|y|1|2|

24

13

8

3

We know that, Mean

i.e. or

or ⇒

Q 2.

Correct4

Incorrect-1

Compute the median from the following table Marks obtained|No. of students| 0-10|2| 10-20|18| 20-30|30| 30-40|45| 40-50|35| 50-60|20| 60-70|6| 70-80|3|

36.55

35.55

40.05

None of these

Marks obtained|No. of students|Cumulative frequency|

0-10|2|2|

10-20|18|20|

20-30|30|50|

30-40|45|95|

40-50|35|130|

50-60|20|150|

60-70|6|156|

70-80|3|159|

Here n = 159, which is odd.

Median number, which is in the class 30-40 (see the row of cumulative frequency 95, which contains 80).

Hence median class is 30-40.

We have l = Lower limit of median class = 30

f = Frequency of median class = 45

C = Total of all frequencies preceding median class = 50

i = Width of class interval of median class = 10

Required median

.

Q 3.

Correct4

Incorrect-1

Consider the following statements (1) Mode can be computed from histogram (2) Median is not independent of change of scale (3) Variance is independent of change of origin and scale Which of these is/are correct

(1), (2) and (3)

Only (2)

Only (1) and (2)

Only (1)

It is obvious.

Q 4.

Correct4

Incorrect-1

The sum of two weights of an Atwood's machine is 16 lbs. The heavier weight descends 64 ft in 4 seconds. The value of weights in lbs are

10, 6

9, 7

8, 8

12, 4

Let the two weights be m_{1} and (lbs.)

.....(i)

Now, the distance moved by the heavier weight in 4 seconds from rest is 64 ft.

If acceleration = f, then ⇒

But ⇒ ⇒ ...(ii)

From (i) and (ii), lbs,

Q 5.

Correct4

Incorrect-1

The following data gives the distribution of height of students Height (in cm)|160|150|152|161|156|154|155| Number of students|12|8|4|4|3|3|7| The median of the distribution is

154

155

160

161

Arranging the data in ascending order of magnitude, we obtain

Height (in cm)|150|152|154|155|156|160|161|

Number Of Students|8|4|3|7|3|12|4|

Cumulative frequency |8|12|15|22|25|37|41|

Here, total number of items is 41, i.e. an odd number. Hence, the median is th i.e. 21^{st} item.

From cumulative frequency table, we find that median i.e. 21^{st} item is 155.

(All items from 16 to 22^{nd} are equal, each = 155)

Q 6.

Correct4

Incorrect-1

A pie chart is to be drawn for representing the following data Items of expenditure|Number of Families| Education|150| Food and clothing|400| House rent|40| Electricity|250| Miscellaneous|160| The value of the central angle for food and clothing would be

90^{0 }

2.8^{0 }

150^{0 }

144^{0} Answer.

Required angle for food and clothing

Q 7.

Correct4

Incorrect-1

What is the standard deviation of the following series Measurements|0-10|10-20|20-30|30-40| Frequency|1|3|4|2|

81

7.6

9

2.26

Class Frequency y_{i },

A = 25f_{i}u_{i}f_{i}u_{i}^{2}|||

0-10|1|5|- 2|- 2|4|

10-20|3|15|- 1|- 3|3|

20-30|4|25| 0|0|0|

30-40|2|35| 1|2|2|

|10|||- 3|9|

⇒ σ = 9

Q 8.

Correct4

Incorrect-1

The S.D. of a variatexis σ. The S.D. of the variate where a, b, c are constant, is

None

Let i.e., i.e. , where

,

⇒ ⇒

⇒ ⇒

⇒ ⇒

Thus, new S.D..

Q 9.

Correct4

Incorrect-1

For the data The Karl Pearson's coefficient is

63

Take

|||||||

4 7 8 3 4|5 8 6 3 5|- 1 2 3 - 2 - 1|0 3 1 - 2 0|1 9 1 4 0|0 9 1 4 0|0 6 3 4 0 |

Total||||||

.

Q 10.

Correct4

Incorrect-1

Three forces are acting at a point in a plane. The angles between and and and are 150° and 120^{0} respectively, then for equilibrium, forces P, Q, R are in the ratio

3:02:01

Clearly, the angle between P and R is

. By Lami's theorem,

Q 11.

Correct4

Incorrect-1

A uniform beam of length 2a rests in equilibrium against a smooth vertical plane and over a smooth peg at a distance h from the plane. If θ be the inclination of the beam to the vertical, then sin^{3}θ is

Let AB be a rod of length 2a and weight W. It rests against a smooth vertical wall at A and over peg C, at a distance h from the wall. The rod is in equilibrium under the following forces :

(i) The weight W at G

(ii) The reaction R at A

(iii) The reaction S at C perpendicular to AB.

Since the rod is in equilibrium. So, the three force are concurrent at O.

In ΔACK, we have, sin θ =

In ΔACO, we have, sin θ =

In ΔAGO, we have sin θ = ;

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