Class 10 Maths Previous Year Questions - Statistics

Ans:
First convert the given table in exclusive form subtract 0.5 from lower limit and add 0.5 to upper limit, so the new table will be:

(i) Modal class is the class with highest frequency, so, it is 19.5 – 24.5. hence, lower limit will be ‘19.5’.
(ii) (a)

N/2 = 365/2 = 182.5
Medium class will be 19.5 – 24.5.
OR
(b) Approx 361 participants are there at Class Interval 44.5-49.5 showing 361 cummulative frequency.
Hence 361 participants are less than 50 year of age who undergo vocational training.
(iii) Empirical relationship between mean, median and mode.
Mode = 3 Median – 2 Mean

Ans: (b)
If each value of observation is increased by 3. then mean is also increased by 3.

Ans: (b)

Here, n/2 = 66/2 = 33
Cumulative frequency just greater than 33 is 37.
So. median class is 10 - 15.Lower limit of median class = 10
Highest frequency is 20 so modal class is 15 - 20.
Sum of the lower limits of the median and modaI class is 10 + 15 = 25 

Ans:

(i) Here, maximum class frequency is 7 and class corresponding to this frequency is 600-800, so the modal class is 600-800.
(ii) Here n/2 = 24/2 = 12
Class whose cumulative frequency just greater than and nearest to n/2 is called median class.
Here, cf. = 13 (>12) and corresponding class 600 - 800 is : median class.
1 = 600. c.f. = 6,f= 7, h = 200
So. the median of the given data is 771.429

Assumed mean a = 1100 and crass size, h = 400 - 200 = 200

So, mean rainfall in the season is 850 mm.
(iii)  Number of sub-division having good rainfall
= 2 + 3 +1 + 1 = 7

Ans: 

Since, 200 = 172 + x ⇒ x = 28
Let the assumed mean, a = 3250 and class size, h = 500

= 3250 - 587.5 = 2,662.5
Mean expenditure = Rs. 2,662.5
Also, we have n/2= 100, which lies in the class interval 2500 - 3000.
Median class is 2500 - 3000.
Here l = 2500, c.f. = 9, f = 28, h = 500

Ans: Table for the given data is as follows:

Now Mean =

Solving for �p: 

∴ p = 96/48= 2

Ans: 

Ans: Let the assumed mean, a = 125 We have the frequency distribution table for the given data as follows :

Hence, mean weight of animals = 123.B kg.

Ans: The frequency distribution table from the given data is as follows:


Hence, the value of f is 16.

Ans: Let the assumed mean, a = 12.5 .-. d
⇒ d = x_i - a = x_i - 12.5  
Now, we have the frequency distribution table as follows:


= 13.9

Ans: We know that
 
Here given l = 65,f₀ = 6, f₁= f, h = 15, f₂ = 8 and mode = 75  
So, from equation (i), we get

Ans: Here the given mode = 240, which lies in interval 200-300.
l = 200, f₀ = 230, f₁= 270, f₂ = x (missing frequency] and h  = 100

Missing frequency, x = 210

Ans: Here, mode of the frequency distribution = 55.
which lies in the class interval 45-60.
∴ Modal class is 45 - 60
Lower limit (l) = 45
Class interval (h) = 15
Also, f ₀ = 15, f ₁ = x and f ₂ = 10

⇒ 10(30 – x – 10) = 225 – 15x
⇒ 300 –10x – 100 = 225 – 15x
⇒ 5x = 25
⇒ x = 5

Ans: The cumulative frequency distribution table is as follows:

Now, we have N = 50

Since, the cumulative frequency just greater than 25 is 27.

∴ The median class is 140 - 145
and also, I = 140, c.f. = 15, f = 12 and h = 5

∴ Median height of the students = 144.16 cm.

Ans: (i) It is clear from the given data, maximum frequency is 33. which lies in 35-40
∴ Modal class is 35-40.

So, modal age of policy holders is 37 years approx.

Ans: (b)
Here the greatest frequency is 7, which lies in the interval 60-80.
So, modal class is 60-80.
Class mark of modal class = upper limit + lower limit / 2
= 60 + 80 / 2 = 70
So, class mark of modal class is 70.

Ans: (d)

Ans: (d)
Here n = 20 ⇒ n /2 = 10
Cumulative frequency just greater than 10 is 15 and corresponding interval is 60-80.
So, median class is 60-80.

Ans: (c)
Median class = 60-80 .
∴ Lower limit of median = 60
Modal class - 60-80 = 120
∴ Lower limit of modal class = 60
So, the sum of lower limit of median and modal class = 60 + 60 = 120

Ans: (a)
From the above  data, we have
l = 60, f = 7, c.f. = 8, h = 20

So, median time (in sec) of the given data = 65.7 sec.

Ans: Given, mean of first n natural numbers is 15.

Ans: In the formula

 
where a is assumed mean and h = class size.

Ans:  The frequency distribution table from the given data can be drawn as :

∴ Mean = 326/40 = 8.15

Ans: From the given data, we have maximum frequency 75. which lies in the interval 20-25.
Modal class is 20-25
So, l = 20, f₀= 30, f₁ = 75,  f₂= 20, h = 5

Mode = 20 + 2.25
= 22.25

Ans: From the given data, we have maximum frequency 12.
which lies in the interval 30-40
Modal class is 30-40
So, l = 30, f₀= 12, f₁ = 7,  f₂= 5, h = 10

Ans: From the given data, we have maximum frequency 12.
which lies in the interval 60-80.
Modal class is 60-80
So, l = 60, f₀= 12, f₁ = 10,  f₂= 6, h = 20

Ans: (b)
We know that Mode = 3 Median - 2 Mean
So, Mode = 3 x  15 - 2 x 14
= 45 - 28 = 17 

Ans: The frequency distribution table for the given data can be drawn as:

= 100 + 26.25
= 126.25

Hence, mean number of wickets is 125.33 and median number of wickets is 126.25.

Ans: The frequency distribution table from the given data is as follows:

Now,      [Given]

Ans: he frequency distribution table from the given data is as follows:

∴
= 50

Ans: Here h = 20
Let us construct the following table far the given data.

We know that Mean =  

Missing frequency, x= 10

Ans: From the given data, we observe that, highest frequency is 20,
which lies in the class-interval 40-50.
∴ l = 40, f₁ =  20, f_o= 12, f₂ = 11, h = 10

Ans: The frequency distribution table for the given data is as follows:

Here, N = 40 ⇒ 31 + f₁ + f₂ = 40
⇒ f₁ + f₂ = 9 ...(i)
Given, median = 32.5, which lies in the class interval 30-40.
So, median class is 30-40.
I = 30, h = 10, f = 12, N = 40 and c.f. of preceding class = f₁ + 14

Ans: The frequency distribution table for the given data is as follows:

Here. N = 100, median = 32, it lies in the Interval 30 - 40.