# 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.

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

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.

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

Q.1 The mean of 200 items is 48 and their S.D. is 3. Find the sum and sum of squares of all the items.

Marks:4

Ans

Q.2 The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

Marks:6

Ans

Squaring (i), we get
(x + y)2 = 132
x2 + y2 + 2xy = 169 …(iii)
From (ii) put value of x2 + y2 in (iii) we get
97 + 2xy = 169
2xy = 72(iv)
Subtracting (iv) from (ii), we get
x2 + y2 + 2xy = 97  72 = 25
(x y)2 = 25
x y = 5 …(v)
From (i) and (v), we get
x = 9 and y = 4 ( when x y = 5)
x = 4 and y = 9 (when x y = 5 )
Thus the two remaining observations are 4 and 9.

Q.3 Find the mean deviation about the median for the following data:

 Class 010 1020 2030 3040 4050 5060 Frequency 7 15 6 16 2 4

Marks:6

Ans

 Class Frequency, f Cumulative Frequency(cf) Mid-pointx ${\mathrm{X}}_{\mathrm{i}}\text{Median}$ ${\mathrm{x}}_{\mathrm{i}}\text{Median}$ $\mathrm{f}{\mathrm{x}}_{\mathrm{i}}\text{Median}$ 0-10 7 7 5 20 20 140 10-20 15 22 15 10 10 150 20-30 6 28 25 0 0 0 30-40 16 44 35 10 10 160 40-50 2 46 45 20 20 40 50-60 4 50 55 30 30 120 50 $\begin{array}{l}\mathrm{f}{\mathrm{x}}_{\mathrm{i}}\text{Median}\\ =610\end{array}$

Q.4 If the coefficient of varitaions of two distributions are 20 and 30 and their standard deviations are 4 and 15, the raio of their arithmetic means are

Marks:1

Ans 2:5.

Q.5  If the S.D. of a variate x is, then the S.D. of ax + b is

Marks:1

Ans

|a|.

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### 1. What do you mean by the term statistics? Name some terminologies associated with statistics.

Statistics refers to the organisation, collection, presentation, interpretation, and analysis of numerical data. Statistics are associated with the following keywords:

• There are two types of class limits: upper and lower limits. A class’s limit is defined as the class’s end value.
• The class interval is defined as the difference between a class’s upper and lower limits.
• Primary data is information gathered directly by the researcher.
• Secondary data is information that was gathered by someone other than the investigator.

### 2. What are the properties of arithmetic mean?

The arithmetic mean has the following properties:

• The AM is unaffected by changes in scale or origin.
• If you add the deviations of a set of values from their arithmetic mean, the result is 0.
• When the sum of a set of values’ square deviations from their arithmetic mean is taken, its value is the smallest.

### 3. What do you understand by the terms mean, median, and mode?

(a) Mean – The ratio of the sum of item values to the total quantity or amount of data is defined as the mean. The following are examples of arithmetic means:

• Mean for unclassified data
• Frequency distribution arithmetic mean
• Mean for combination
• Mean for classified data
• Arithmetic mean for weights

(b) Median – The median is the middle value of a data set when it is arranged in ascending or descending order.

(c) Mode – This is the value of the point in the data set where it is highly concentrated.

### 4. Describe the following terms:

(i) Cumulative frequency distribution

Ans: The cumulative frequency is the sum of the frequencies of the first, second, third, and so on, which results in the final frequency. Class frequencies are required for this frequency distribution.

(ii) Discrete frequency distribution

Ans: In this type of frequency distribution, the data is presented in such a way that the unit measurements are clearly visible.

(iii) Continuous frequency distribution

Ans: A continuous frequency distribution is a type of frequency distribution in which groups of classes are not measurable.