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)
New Mean = Old Mean + 3
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
∴ Median = l + n2 - c.f.f × h

= 600 + 12 - 67 × 200

= 600 + 67 × 200

= 771.429

Assumed mean a = 1100 and class size, h = 400 - 200 = 200
∴ Mean = a + h∑ f_i [∑ f_i u_i]

= 1100 + 20024 × (-30)

= 1100 - 600024

= 850

Ans: 

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

Mean( x̄ ) = a + h × 1n ∑ f_i u_i

= 3250 + 500 × 1200 × (-235)

= 3250 - 587.5 = 2,662.5

Mean expenditure = ₹ 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. = 97, f = 28, h = 500

Median = l + n2 - c.f.f × h

= 2500 + 100 - 9728 × 500

= 2500 + 328 × 500

= 2553.57

∴ Median expenditure = ₹ 2553.57

Ans: (c)
Step 1: Identify the Class Intervals
The table provided shows cumulative frequencies for marks below certain values:

• Marks below 10: 3 students
• Marks below 20: 12 students
• Marks below 30: 27 students
• Marks below 40: 57 students
• Marks below 50: 75 students
• Marks below 60: 80 students

To find the frequencies for each class, we calculate the difference between consecutive cumulative frequencies:

1. 10−20: 12−3=9
2. 20−30: 27−12=15
3. $330\; -\; 40$30−40: $57\; -\; 27\; =\; 30$57−27=30
4. $40\; -\; 50$40−50: 75−57=18
5. 50−60: $80\; -\; 75\; =\; 5$80−75=5

So, the frequencies for each interval are:

• 10−20: 9
• $20\; -\; 30$20−30: 15
• $30\; -\; 40$30−40: 30
• 40−50: 18
• $5$50−60: 5

Step 2: Determine the Modal Class
The modal class is the class interval with the highest frequency. From the calculated frequencies, the highest frequency is 30, which corresponds to the class interval $30\; -\; 40$30−40.
The modal class is: (c) 30−40.

Ans: Table for the given data is as follows:

Now Mean =

10.8 = (2 × 3) + (6 × p) + (10 × 5) + (14 × 8) + (18 × 2)3 + p + 5 + 8 + 2

Solving for p:

10.8 = 6 + 6p + 50 + 112 + 3618 + p

10.8 = 204 + 6p18 + p

(10.8)(18 + p) = 204 + 6p

194.4 + 10.8p = 204 + 6p

4.8p = 9.6

p = 9.64.8

p = 2

Ans: 

Midpoint = Lower Limit + Upper Limit2

For the given class intervals:

• Midpoint of 0-10: 0 + 102 = 5
• Midpoint of 10-20: 10 + 202 = 15
• Midpoint of 20-30: 20 + 302 = 25
• Midpoint of 30-40: 30 + 402 = 35
• Midpoint of 40-50: 40 + 502 = 45
• Midpoint of 50-60: 50 + 602 = 55

Now, we'll multiply each midpoint by its corresponding frequency, sum up these products, and divide by the total frequency to find the mean:

Mean = ∑ (Midpoint × Frequency)Total Frequency

So, the mean is:

Mean = (5 × 12) + (15 × 18) + (25 × 27) + (35 × 20) + (45 × 17) + (55 × 6)

= 60 + 270 + 675 + 700 + 765 + 330

Mean = 2800100

Mean = 28

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

∴ Mean ( x̄ ) = a + 1N ∑ f_i d_i

= 125 + 150 × (-60)

= 125 - 6050

= 125 - 1.2 = 123.8

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

∴ Mean ( x̄ ) = ∑ f_i x_i∑ f_i

⇒ 25 = 940 + 35f44 + f     [∵ Given, mean = 25]

⇒ 25(44 + f) = 940 + 35f

⇒ 1100 + 25f = 940 + 35f

⇒ 10f = 160

⇒ f = 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:

∴ Mean ( x̄ ) = a + 1N ∑ f_i d_i

= 12.5 + 7050

= 12.5 + 1.4

Ans: We know that

Mode = l + f₁ - f₀2f₁ - f₀ - f₂ × h ... (i)

Here given l = 65, f₀ = 6, f₁ = f, h = 15, f₂ = 8 and mode = 75

So, from equation (i), we get

75 = 65 + f - 62f - 6 - 8 × 15

75 = 65 + f - 62f - 14 × 15

75 - 65 = (f - 6) × 152f - 14

(2f - 14) × 10 = 15f - 90

⇒ 20f - 15f = -90 + 140

⇒ 5f = 50

∴ f = 10

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

∴ Mode = l + f₁ - f₀2f₁ - f₀ - f₂ × h

240 = 200 + 270 - 2302 × 270 - 230 - x × 100

240 = 200 + 40310 - x × 100

⇒ 240 - 200 = 4000310 - x

⇒ 40 = 4000310 - x

⇒ 310 - x = 100

⇒ x = 310 - 100 = 210

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

Mode = l + f₀ - f₁2f₀ - f₁ - f₂ × h

⇒ 55 = 45 + 15 - x30 - x - 10 × 15

⇒ 55 - 45 = 15(15 - x)30 - x - 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

⇒ N2 = 502 = 25

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

∴ The median class is 140 - 145

And also, l = 140, c.f. = 15, f = 12 and h = 5

∴ Median = l + N2 - c.f.f × h

= 140 + 25 - 1512 × 5

= 140 + 1012 × 5

= 140 + 4.16 = 144.16

∴ 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, l = 35, f₀ = 21, f₁ = 33, f₂ = 11, and h = 5

∴ Mode = l + f₁ - f₀2f₁ - f₀ - f₂ × h

So, modal age of policy holders

= 35 + 33 - 212 × 33 - 21 - 11 × 5

= 35 + 1234 × 5

= 35 + 6034

= 35 + 1.76

= 36.76 (approx)

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)

h = 20, l = 60, f₁ = 7, f₀ = 3, f₂ = 5

∴ Mode = l + f₁ - f₀2f₁ - f₀ - f₂ × h

= 60 + 7 - 32 × 7 - 3 - 5 × 20

= 60 + 46 × 20

= 60 + 13.33

Mode = 73.33

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

∴ Median = l + n2 - c.f.f × h

= 60 + 202 - 87 × 20

= 60 + 10 - 87 × 20

= 60 + 407

= 60 + 5.714

= 65.71 (approx)

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

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

⇒ 1 + 2 + 3 + ..... + nn = 15

⇒ 1 + 2 + 3 + ..... + n = 15n

⇒ n(n + 1)2 = 15n

⇒ n² + n = 30n

⇒ n² - 29n = 0

⇒ n(n - 29) = 0

⇒ n = 29 [n ≠ 0]

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 = l + f₁ - f₀2f₁ - f₀ - f₂ × h

= 20 + 75 - 302(75) - 30 - 20 × 5

= 20 + 45100 × 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

Mode = l + f₁ - f₀2f₁ - f₀ - f₂ × h

= 30 + 12 - 72 × 12 - 7 - 5 × 10

= 30 + 524 - 12 × 10

= 30 + 5012

= 30 + 4.17

= 34.17

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

Mode = l + f₁ - f₀2f₁ - f₀ - f₂ × h

= 60 + 12 - 102 × 12 - 10 - 6 × 20

= 60 + 224 - 16 × 20

= 60 + 28 × 20

= 60 + 5

= 65

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:

Mean = ∑ f_i x_i∑ f_i = 564045 = 125.33

Here, N2 = 452 = 22.5

∴ Median class is 100-140.

Also, l = 100, c.f. = 12, f = 16, h = 40

So, Median = l + N2 - c.f.f × h

= 100 + 22.5 - 1216 × 40

= 100 + 10.516 × 40

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

Ans: 

Here, Σf_i = 41 + p
and Σf_ix_i = 303 + 9p

We know, Mean = ∑ f_i x_i∑ f_i

But, Mean = 7.5 [Given]

∴ 7.5 = 303 + 9p41 + p

⇒ 303 + 9p = 307.5 + 7.5p

⇒ 1.5p = 4.5

⇒ p = 3

Hence, the value of p is 3.

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

Now, mean = ∑ f_i x_i∑ f_i = 53 [Given]

∴ 3340 + 70k72 + k = 53

⇒ 3340 + 70k = 3816 + 53k

⇒ 70k - 53k = 3816 - 3340

⇒ 17k = 476

⇒ k = 28

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 = ∑ f_i x_i∑ f_i

⇒ 62.8 = 2640 + 50x40 + x

⇒ 62.8(40 + x) = 2640 + 50x

⇒ 2512 + 62.8x = 2640 + 50x

⇒ 62.8x - 50x = 2640 - 2512

⇒ 12.8x = 128 ⇒ x = 12812.8

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

Mode = l + f₁ - f₀2f₁ - f₀ - f₂ × h

= 40 + 20 - 1240 - 12 - 11 × 10

= 40 + 8017

= 40 + 4.7 = 44.7

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

Now, median = l + N2 - c.f.f × h

⇒ 32.5 = 30 + 20 - (f₁ + 14)12 × 10

⇒ 2.5 = 6 - f₁12 × 10

⇒ 6 - f₁ = 3 ⇒ f₁ = 3

From (i), f₂ = 9 - 3 = 6

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

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

∴ Median = l + N2 - c.f.f × h

⇒ 32 = 30 + 50 - (35 + x)30 × 10

⇒ 32 - 30 = 15 - x3

⇒ 15 - x = 6

⇒ x = 9

Also, 75 + x + y = 100

⇒ 75 + 9 + y = 100

⇒ y = 100 - 84 = 16