Important Questions Class 11 Maths Chapter 15

Important Questions for CBSE Class 11 Maths Chapter 15 – Statistics

Statistics is the collection, organisation, and presentation of numerical data for analysis and interpretation. Statistics in Class 11 Mathematics is one of the most important chapters. Most of the concepts are likely to be introduced to the students for the first time. Hence,  it is critical for them to clearly understand these fundamental concepts from the start.

Extramarks’ Important Questions for Class 11 Mathematics Chapter 15 presents important topics in a question-answer format. These questions are compiled by subject matter experts by referring to NCERT books and past years’ question papers.

CBSE Class 11 Maths Chapter-15 Important Questions

Study Important Questions for Class 11 Mathematics Chapter 15 – Statistics

Some of the Important Questions for Class 11 Mathematics Chapter 15 each for 1, 4, and 6 marks are given below. The complete set of questions can be accessed by clicking on the link provided.

1 Mark Answers and Questions

Q1. Find the mean of the following data  3,6,11,12,18 

A1. Mean = Sum of observation / Total no of observations

=50/5

=10

Therefore, the mean of the given data is 10.

Q2. Find the range of the following series  6,7,10,12,13,4,8,12 

A2. Range =  Maximum value – Minimum value

=13−4

=9

Therefore, the range of the given series is 9.

Q3. Write the formula for mean deviation from the median.

A3. The formula for mean deviation from the median is: MD.(M)=fixiMfi=1nfixi-M

4 Marks Answers and Questions

Q1. Find the mean deviation about the mean for the following data.

xi:10,30,50,70,90

fi:4,24,28,16,8

A1. To calculate the mean, we require fixed values then for mean deviation, it requires

∣xi−x¯∣ values and fi|xi−x¯| values.

n=∑fi=80

σd∑fixi=4000

x¯=∑fixin=4000/80=50

The mean deviation about the mean is,

MD(x¯)=∑fi|xi−x¯|n=1280/80=16

Q2. Demonstrate that the standard deviation is independent of the origin but dependent on the scale.

A2. To change the scale and origin, use the transformation u=ax+b.

Now  u=ax+b

=∑u=Σ(ax+b)=a∑x+b.n

Also σu2=∑(u−u¯)2n

=∑(ax+b−ax¯−b)2n

=∑a2(x−x¯)2n

=a2Σ(x−x¯)2n

=a2σx2

σ2u=a2σ2u

=σu=∣a∣σx

Both σu and σx are positive which shows that the standard deviation is independent of the choice of origin, but depends on the scale.

Q3. An examination of monthly wages reveals that workers in two firms A and B, both in the same industry, have the following result. Determine the mean deviation from the median.

The total number of wages earned is 586 and 648 in Firm A and Firm B respectively.

The average monthly wage is Rs 5253.

Ans. The number of employees at firm A is 586.

The average monthly salary is Rs 5253.

Total earnings = Rs5253586

=Rs3078258

Total wages for firm B = Rs253648

=Rs3403944

As a result, firm B pays out a certain amount of monthly wages.

Q4. Find the real values of x and y if (x−iy)(3+5i) is the conjugate of −6−24i

A4. (x−iy)(3+5i)=−6+24i

Expand brackets,

3x+5xi−3yi−5yi2=−6+24i

Group terms,

(3x+5y)+(5x−3y)i=−6+24i

3x+5y=−6

5x−3y=24

x=3

y=−3

Q5. If |z1|=|z2|=1, prove that ∣∣∣1z1+1z2∣∣∣=|z1+z2|

A5. Given that, |z1|=|z2|=1

⇒|z1|2=|z2|2=1

⇒z1z1¯¯¯¯¯=1

z1¯¯¯¯¯=1z1→(1)

z2z2¯¯¯¯¯=1

z2¯¯¯¯¯=1z2→(2)   [∵zz¯=|z|2

∣∣∣1z1+1z2∣∣∣=|z1¯¯¯¯¯+z2¯¯¯¯¯|

=∣∣z1+z2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣

=|z1+z2|   [∵|z¯|=|z|

Q6. Find the modulus i25+(1+3i)3

A6. Given, i25+(1+3i)3

=(i4)6⋅i+1+27i3+3(1)(3i)(1+3i)

=i+(1−27i+9i+27i2)

=i+1−18i−27

=−26−17i

∣∣i25+(1+3i)3∣∣=|−26−17i|

=(−26)2+(−17)2−−−−−−−−−−−−−√

=(−26)2+(−17)2−−−−−−−−−−−−−√

=676+289−−−−−−−−√

=965−−−√

Q7. The average and standard deviation of six observations are eight and four, respectively. Find the new mean and standard deviation of the resulting observations after multiplying each observation by 3.

A7. Let xi,x2…..x6 represent the six given observations.

Then, x¯=8 and σ=4

x¯=∑xin=8=x1+x2+……+x66

x1+x2+……x6=48

Also σ2=∑x21n−(x¯)2

Substitute the values,

=42=x21+x22……+x266−(8)2

=x21+x22+……x26

=6×(16+64)=480

As each observation is multiplied by  3 , new observations are,

3×1,3×2,……3×6

Then, X¯=3×1+3×2+….3×66

=3(x1+x2+….x6)6

=3×486

=24

Let σ1 be the new standard deviation, then

σ21=(3×1)2+(3×2)2+…..+(3×6)26−(X¯)2

=9(x21+x22+….x26)6−(24)2

=9×4806−576

=720−576

=144

σ1=12

Q8. If (x+iy)3=u+iv then show that ux+vy=4(x2−y2)

A8. (x+iy)3=u+iv

x3+(iy)3+3×2(iy)+3⋅x(iy)2=u+iv

x3−iy3+3x2yi−3xy2=u+iv

x3−3xy2+(3x2y−y3)i=u+iv

x(x2−3y2)+y(3×2−y2)i=u+iv

x(x2−3y2)=u,y(3×2−y2)=v

x2−3y2=ux     ∣3×2−y2=vy

4×2−4y2=ux+vy

4(x2−y2)=ux+vy

Hence proved.

6 Marks Answers and Questions

Q1. A student calculated the mean and standard deviation of 100 observations as 40 and 5.1, respectively, after mistaking one observation for 50 instead of 40. What is the appropriate mean and standard deviation?

A1. Given that, n=100

Incorrect mean x¯=40,

Incorrect S.D (σ)=5.1

As x¯=∑xin

40=∑xi100⇒∑xi=4000

⇒ Incorrect sum of observation =4000

⇒ Correct sum of observations =4000−50+40

=3990

Now, Correct mean =3990100=39.9

Also, σ=1n∑x2i−(x¯)2−−−−−−−−−−−−√

Use incorrect values,

5.1=1100∑x2i−(40)2−−−−−−−−−−−−−−√

⇒26.01=[1100∑x2i−1600]

=162601

Incorrect ∑x2i=162601

Correct ∑x2i=162601−(50)2+(40)2

=162601−2500+1600=161701

Correct σ=1100 correct ∑x2i−( correct x¯)2−−−−−−−−−−−−−−−−−−−−−−−−−−√

=1100(161701)−(39.9)2−−−−−−−−−−−−−−−−−−√

=1617.01−1592.01−−−−−−−−−−−−−−√

=25−−√=5

Therefore, the correct mean is  39.9  and the correct standard deviation is  5.

Q2. Convert into polar form z=i−1cosπ3+iSinπ3

A2. Given that, z=i−1cosπ3+iSinπ3

Substitute the known values,

z=i−112+3–√2i

=2(i−1)1+3–√i×1−3–√i1−3–√i

z=3–√−12+3–√+12i

r=|z|=(3–√−12)2+(3–√+12)2

r=2

Let α be the acute ∠s,

tanα=∣∣∣∣∣3–√+123–√−12∣∣∣∣∣

=∣∣∣∣∣∣3–√(1+13–√)3–√(1−13–√)∣∣∣∣∣∣

=∣∣∣∣∣tanπ4+tanπ61−tanπ4tanπ6∣∣∣∣∣

tanα=∣∣tan(π4+π6)∣∣

α=π4+π6=5π12

z=2(Cos5π12+iSin5π12)

Q3. If α and β are different complex numbers with |β|=1. Then find ∣∣∣β−α1−α¯β∣∣∣

A3. ∣∣∣β−α1−α¯β∣∣∣2=(β−α1−α¯β)(β−α¯¯¯¯¯¯¯¯¯¯¯¯1−α¯β)[∵|z|2=zz¯

=(β−α1−α¯β)(β¯−α¯1−αβ¯)

=(ββ¯−βα¯−αβ¯+αα¯1−αβ−αβ+αα¯ββ¯)

=|β|2−βα¯−αβ¯+|α|21−αβ¯−α¯β+|α|2|β|2}

1−βα¯−αβ¯+|α|2

1−αβ¯¯−α¯β¯¯¯¯¯¯β+|α|2

∵ |β|=1

=1

∣∣∣β−α1−αβ∣∣∣=1–√

∣∣∣β−α1−α⃗ β∣∣∣= 1