Unit Test (Solutions): Statistics | Mathematics (Maths) Class 10 PDF Download

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Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: If the mean of first n natural numbers is 3n/5, then the value of n is: (1 Mark)

(a) 3
(b) 4
(c) 5
(d) 6

Ans: (c)

Sum of natural numbers = n(n + 1)/2
Given, mean = 3n/5
Mean = sum of natural numbers/n
3n/5 = n(n + 1)/2n
3n/5 = (n + 1)/2
6n = 5n + 5
n = 5

Q2: The class interval of a given observation is 10 to 15, then the class mark for this interval will be: (1 Mark)

(a) 11.5
(b) 12.5
(c) 12
(d) 14

Ans: (b)

Class mark = (Upper limit + Lower limit)/2
= (15 + 10)/2
= 25/2
= 12.5

Q3: If the sum of frequencies is 24, then the value of x in the observation: x, 5,6,1,2, will be: (1 Mark)

(a) 4
(b) 6
(c) 8
(d) 10

Ans: (d)

Given,
∑fi = 24
∑fi = x + 5 + 6 + 1 + 2 = 14 + x
24 = 14 + x
x = 24 – 14 = 10

Q4: The ________ of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class. (1 Mark)

Ans: Cumulative frequency

The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.

Q5: The method used to find the mean of a given data is(are): (1 Mark)

(a) direct method
(b) assumed mean method
(c) step deviation method
(d) all the above

Ans: (d)

The mean for a given data can be calculated using either of the following methods:
i) Direct method
ii) Assumed mean method
iii) Step deviation method

Q6: If the mode of the following data is 7, then find the value of k. (2 Marks)
2, 4, 6, 7, 5, 6, 10, 6, 7, 2k + 1, 9, 7, 13

Ans: There are two entries which occur 3 times each which are 6 and 7.
For 7 to be mode
2k + 1 = 7
k = 6/2 = 3

Q7: The mean of 6 numbers is 16 with the removal of a number the mean of remaining numbers is 17. Find the removed number.  (2 Marks)

Ans: Let x_d be the sixth number to be remove and x₁ x₂, …,x₅ be remaining 5 numbers. 

Q8: The mean of following frequency distribution is 70. Find the missing frequency x. (2 Marks)

Ans: 

Q9: Find the mode of the following distribution of marks obtained by the students in an examination:
Given the mean of the above distribution is 53, using emperical relationship estimate the value of its median.  (3 Marks)
Ans: Modal class is 60-80 (Class corresponding to maximum frequency)So, the mode marks is 68.
Empirical relationship between the three measures of central tendencies is:
3 Median = Mode + 2 mean
3 Median = 68 + 2 × 53
Median = 58 marks

Q10: Find the missing frequecny (x) of the following data if its mode is ₹240.  (3 Marks)
Ans: Here mode is given to be ₹ 240, which lies between 200 and 300. Therefore, modal class is 200 – 300
We have l = 200, f₁ = 270, f₀ = 230, f₂ = x12400 – 40x = 4000
40x = 12400 – 4000
40x = 8400
x = 210

Q11: Consider the following distribution of daily wages of 50 workers of a factory. (3 Marks)
Find the mean daily wages of the workers of the factory by using an appropriate method.

Ans:  Find the midpoint of the given interval using the formula.

Midpoint (x_i) = (upper limit + lower limit)/2

In this case, the value of mid-point (x_i) is very large, so let us assume the mean value, a = 550.

Class interval (h) = 20

So, u_i = (x_i – a)/h

u_i = (x_i – 550)/20

Substitute and find the values as follows:So, the formula to find out the mean is:Thus, mean daily wage of the workers = Rs. 545.20

Q12: The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f₁ and f₂. (3 Marks)

Ans: 

Now 50 = 5 + f₁ + 10 + f₂ + 7 + 8
⇒ 50 = 30 + f₁ + f₂
⇒ f₁ + f₂ = 20 ……….(i)

3140 = 2060 + 30f₁ + 70f₂
3f₁ + 7f₂ = 108
Solving (i) and (ii), we get
f₁ = 8 and f₂ = 12

Q13: Find the mean, mode and median of the following data: 

Ans: We prepare the table which serves all three purposes of getting mean, median and mode as given below:
(i) Mean: Let assumed mean (A) be 35.

(ii) Mode: As maximum frequency is 15.

∴ Mode classes is 30-40.

Thus f₁ = 15, f₀ = 10, f₂ = 7 

(iii) Median: Here N = 50 ⇒ N/2 = 25

The cumulative frequency just greater than 25 is 36 and the corresponding class is 30-40.

∴ Median class = 30-40, l = 30, cf = 21, f = 15