WBBSE Class 9 Maths Statistics Chapter 1 Statistics Multiple Choice Questions

WBBSE Class 9 Maths Statistics Chapter 1 Statistics Multiple Choice Questions

Example 1. Which one of the following is a graphical representation of statistical data?

1. Line-graph
2. Raw data
3. Cumulative frequency
4. Frequency distribution

Solution: Line graph is a graphical representation of statistical data.

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∴ So the correct answer is 1. Line-graph

Example 2. The range of the data 12, 25, 15, 18, 17, 20, 22, 26, 6, 16, 11, 8, 19, 10, 30, 20, 32 is

1. 10
2. 15
3. 18
4. 26

Solution: The range = 32 – 6 = 26

∴ The correct answer is 4. 26

Example 3. The class size of the classes 1-5, 6-10 is

1. 4
2. 5
3. 4.5
4. 5.5

Solution: The difference (d) between classes 1-5, and 6-10, is 1

The lower class boundary of class 1-5 is (1- \(\frac{1}{2}\)) or 0.5 and the upper-class boundary is (5 + \(\frac{1}{2}\)) or 5.5

The class size = upper class boundary – lower class boundary = 5.5 – 0.5 = 5

∴ The correct answer is 2. 5

Example 4. In a frequency distribution table, the midpoints of the classes are 15, 20, 25, and 30, respectively. The class having a midpoint as 20 is

1. 12.5 – 17.5
2. 17.5 – 22.5
3. 18.5 – 21.5
4. 19.5 – 20.5

Solution: The difference between two consecutive midpoints

= 20 – 15 = 25 – 20 = 30 – 25 = 5

So, the class size = 5

The midpoint of class (12.5 – 17.5) = \(\frac{12 \cdot 5+17 \cdot 5}{2}=\frac{30}{2}=15\)

and class-length = 17.5 – 12.5 = 5

The midpoint of class (17.5 – 22.5) = \(\frac{17 \cdot 5+22 \cdot 5}{2}=\frac{40}{2}=20\)

and class length = 22.5 – 17.5 = 5

The midpoint of class (18.5 – 21.5) = \(=\frac{18 \cdot 5+21 \cdot 5}{2}=20\)

and class length = 21.5 – 18.5 = 3

The midpoint of class (19.5 – 20.5) = \(=\frac{18 \cdot 5+21 \cdot 5}{2}=20\)

and class length = 20.5 – 19.5 = 1

∴ The required class is 17.5 – 22.5

∴ The correct answer is 2. 17.5 – 22.5

Example 5. In a frequency distribution table, the midpoints of a class is 10 and the class size of each class is 6; the lower limit of the class is

1. 6
2. 7
3. 8
4. 12

Solution: Let, lower limit of given class is x;

As class size is 6 then upper limit is (x+6)

The midpoint = \(\frac{x+x+6}{2}=\frac{2 x+6}{2}=x+3\)

As per question, x + 3 = 10

⇒ x = 10 – 37 = 7

∴ The lower limit is 7

∴ The correct answer is 2. 7

Example 6. Each of area of each of the rectangles of a histogram is proportional to

1. The mid-point of that class
2. The class size of that class
3. The frequency of that class
4. The cumulative frequency of that class

Solution: Which increasing frequency the area of each of the rectangles of a histogram will also increase.

So the area of a rectangle is proportional to the frequency.

∴ The correct answer is 3. The frequency of that class

Example 7. A frequency polygon is drawn by the frequency of the class and

1. The upper limit of the class
2. The lower limit of the class
3. Mid-value of the class
4. Any value of the class

Solution: A frequency polygon is drawn by the frequency of the class and mid value of the class.

∴ The correct answer is 3. Mid-value of the class

Example 8. To draw a histogram, the class boundaries are taken

1. Along Y-axis
2. Along X-axis
3. Along X-axis and Y-axis both
4. In between X-axis and Y-axis

Solution: To draw a histogram, the class boundaries are taken along X-axis.

∴ The correct answer is 2. Along X-axis

Example 9. In case of drawing a histogram, the base of the rectangle of each class is

1. Frequency
2. Class-boundary
3. Range
4. Class-size

Solution: In case of drawing a histogram the length of base of a rectangle of each class is class size.

class- the size of each class is 5.

∴ The correct answer is 4. Class-size

Example 10. A histogram is the graphical representation of grouped data whose class boundary and frequency are taken respectively.

1. Along the vertical axis and the horizontal axis
2. Only along the vertical axis
3. Only along the horizontal axis
4. Only the horizontal axis and vertical axis

Solution: A histogram is the graphical representation of grouped data whose class boundary and frequency are taken respectively only the horizontal axis and vertical axis.

∴ The correct answer is 4. Only the horizontal axis and vertical axis

Example 11.

The frequency density of the third class of the above frequency distribution table is

1. 1.5
2. 0.55 (approx)
3. 0.36
4. 2.8 (approx)

Solution: The difference between two consecutive class is 1

∴ Lower boundary of third class is \(\left(21-\frac{1}{2}\right)\) or 20.5

and upper-boundary of third class is \(\left(30+\frac{1}{2}\right)\) or 30.5

∴ class-size = 30.5 – 20.5 = 10

The frequency density of third class = \(\frac{\text { class frequency }}{\text { class size }}=\frac{18}{10}\) = 1.8

∴ The correct answer is 1. 1.5

Example 12. The upper-class boundary of the fifth class of the classes 30-39, 40-49,….. is

1. 69.5
2. 79.5
3. 69
4. 79

Solution: Classes are 30-39, 40-49, 50-59, 60 – 69, 70-79,……

The fifth class is 70 – 79

The difference between the two consecutive class is 1

∴ The upper-class boundary of the fifth class is \(\left(79+\frac{1}{2}\right)\) or, 79.5

∴ The correct answer is 2. 79.5

Example 13. If lower class boundary and mid value of a class are a and m respectively, then the upper class boundary will be

1. a + 2m
2. a – 2m
3. 2m – a
4. 2a – m

Solution: Let upper-class boundary of the given class is b

As lower class boundaries a

So the mid value (m) = \(\frac{a+b}{2}\)

or, a + b = 2m or, b = 2m – a

∴ The correct answer is 3. 2m -a

∴ Then the upper class boundary will be 2m -a