NCERT Solutions For Class 10 Maths Chapter 14: Statistics

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NCERT Solutions for class 10 Maths Chapter 14 Statistics are provided in the article below. Some of the important topics of Statistics chapter include:

Expected no of questions: 2 to 3 questions of total 5 marks

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NCERT Solutions for Class 10 Mathematics Chapter 14

NCERT Solutions for Class 10 Maths Chapter 14 Statistics is given below.

NCERT Solution class 10 mathematics chapter 14 statistics

Class 10 Mathematics Chapter 14 Statistics – Important Topics

Study of collection, presentation, organisation and analysis of data is known as Statistics.

Important formulas in Statistics include:

Mean x = ∑x / n

Median [if n is odd] = (n+1 /2)

Median [if n is even] = [(n/2)th + (n+1 /2)] / 2

Standard Deviation σ = √∑ni=1(xi-μμ)² /n

Variance σ² = ∑ni=1(xi-μμ)² /n

Average of a given set of numbers is known as the mean.
Middle Value in a given set of data is known as median.
Most Occurring item in a given set of data is known as mode.
A measure which gives an approximate idea about data is known as variance. It is used to calculate standard deviation of a data set.
We can find how far a value is from the mean by the help of Standard Deviation. It can be found by taking the square root of Variance.

NCERT Solutions for Class 10 Maths Chapter 14 Exercises

NCERT Solutions for Class 10 Mathematics Chapter 14 Statistics Exercises are given below.

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CBSE X Related Questions

1.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
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2.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
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3.
The natural number 1 is :
- a prime number.
- a composite number.
- prime as well as composite.
- neither prime nor composite.
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4.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals
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- \(10.5\) cm
- \(3.5\) cm
- \(\frac{14}{3}\) cm
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5.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :
- \( 130^\circ \)
- \( 50^\circ \)
- \( 120^\circ \)
- \( 90^\circ \)
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6.
The HCF of 960 and 432 is :
- 48
- 54
- 72
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