## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D.

**Other Exercises**

- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H

**Question 1.**

**Find the arithmetic mean of**

**(i) The first eight natural numbers.**

**(ii) The first ten odd numbers.**

**(iii) The first five prime numbers.**

**(iv) The first six even numbers.**

**(v) The first seven multiples of 5.**

**(vi) All the factors of 20.**

**Solution:**

(i) First eight natural numbers are 1, 2, 3, 4, 5, 6, 7, 8

∴ Mean = \(\frac { 1+2+3+4+5+6+7+8 }{ 8 } \) = \(\frac { 36 }{ 8 } \) = \(\frac { 9 }{ 2 } \) = 4.5

(ii) First ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

∴ Mean = \(\frac { 1+3+5+7+9+11+13+15+17+179 }{ 10 } \) = \(\frac { 100 }{ 10 } \) = 10

(iii) First five prime numbers are 2, 3, 5, 7, 11.

∴ Mean = \(\frac { 2+3+5+7+11 }{ 5 } \) = \(\frac { 28 }{ 5 } \) = 5.6

(iv) First six even numbers are 2, 4, 6, 8, 10, 12

∴ Mean = \(\frac { 2+4+6+8+10+12 }{ 10 } \) = \(\frac { 42 }{ 6 } \) = 7

(v) First seven multiples of 5 are 5, 10, 15, 20, 25, 30, 35

∴ Mean = \(\frac { 5+10+15+20+25+30+35 }{ 7 } \) = \(\frac { 140 }{ 7 } \) = 20

(vi) All the factors of 20 are, 1, 2, 4, 5, 10, 20

∴ Mean = \(\frac { 1+2+4+5+10+20 }{ 6 } \) = \(\frac { 42 }{ 6 } \) = 7

**Question 2.**

**The number of children in 10 families of a locality are 2, 4, 3, 4, 2, 0, 3, 5, 1, 6. Find the mean number of children per family.**

**Solution:**

No. of families (n) = 10

Sum of children (∑x1) = 2 + 4 + 3 + 4 + 2 + 0 + 3 + 5 + 1 + 6 = 30

∴ Mean \(\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 30 }{ 10 } =3\)

**Question 3.**

**The following are the number of books issued in a school library during a week : 105, 216, 322, 167, 273, 405 and 346. Find the average number of books issued per day.**

**Solution:**

Here number of days (n) = 7

Number of books (∑x1) = 105 + 216 + 322 + 167 + 273 + 405 + 346 = 1834

∴ Mean \(\left( \overline { x } \right) =\frac { \sum { x1 } }{ n } =\frac { 1864 }{ 7 } =262 books\)

**Question 4.**

**The daily minimum temperature recorded (in degree F) at a place during a week was as under:**

**Find the mean temperature.**

**Solution:**

Number of days (n) = 6

Sum of temperature (∑x) = 35.5 + 30.8 + 27.3 + 32.1 + 23.8 + 29.9 = 179.4°F

∴ Mean temperature = \(\frac { \sum { x } }{ n } =\frac { 179.4 }{ 6 } \)=29.9°F

**Question 5.**

**The percentages of marks obtained by 12 students of a class in Mathematics are 64, 36, 47 ,23, 0, 19, 81, 93, 72, 35, 3, 1.****Find the mean percentage of marks.**

**Solution:**

Number of students (n) = 12

Sum of marks (∑x) = 64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1 = 474

= \(\frac { \sum { x1 } }{ n } =\frac { 474 }{ 12 } \) = 39.5

**Question 6.**

**If the arithmetic mean of 7, 9, 11, 13, x, 21 is 13 find the value of x**

**Mean percentage of marks \(\left( \overline { x } \right) \)**

**Solution:**

Here, n = 6

and arithmetic mean =13

Total sum = 13 x 6 = 78.

But sum of 7 + 9+ 11 + 13 + 21=61

Value of x = 78 – 61 = 17

Hence x = 17 Ans.

**Question 7.**

**The mean of 24 numbers is 35. If 3 is added to each number, what will be the new mean?**

**Solution:**

Let x1, x2, x3, … x24 be the 24 numbers

\(\frac { x1+x2+x3+…..+x24 }{ 24 } =35\)

=> x1 + x2 + x3 +….+ x24 = 35 x 24

**Question 8.**

**The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers, what will the new mean ?**

**Solution:**

Let x1 + x2 + x3……..x20 be the 20 numbers

**Question 9.**

**The mean of 15 numbers is 27. If each number is multiplied by 4, what will be the mean of the new numbers.**

**Solution:**

Let x1, x2, x3 … x15 be the numbers

Mean = \(\frac { x1+x2+x3+…..+x15 }{ 15 } = 27\)

**Question 10.**

**The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of new numbers.**

**Solution:**

Let x1, x2, x3 … x12 be the numbers

Mean = \(\frac { x1+x2+x3+…..+x12 }{ 12 }\) = 40

=> x1+x2+x3+….+x12 = 40 X 12 =480

Now,new numbers are \(\frac { x1 }{ 8 } ,\frac { x2 }{ 8 } ,\frac { x3 }{ 8 } ,..\frac { x12 }{ 8 } \)

Mean = \(\cfrac { \frac { x1 }{ 8 } +\frac { x2 }{ 8 } +\frac { x3 }{ 8 } +…..+\frac { x12 }{ 8 } }{ 12 } \)

= \(\frac { 1 }{ 8 } \cfrac { \left( x1+x2+x3+….x12 \right) }{ 12 } \)

= \(\frac { 480 }{ 8X12 } \) = 5

Mean of new numbers =5

**Question 11.**

**The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of the new set of 20 numbers.**

**Solution:**

Let x1, x2, x3,…..x20 are the numbers

Mean = \(\frac { x1+x2+x3+…x20 }{ 20 }\) = 18

=> x1 + x2 + x3 +….+ x20 = 18 X 20 =360

Mean of the new numbers = \(\frac { \left( x1+3 \right) +\left( x2+3 \right) +\left( x3+3 \right) +….+\left( x10+3 \right) +x11+x12+….+x20 }{ 20 } \)

= \(\frac { x1+x2+x3+….+x20+3X10 }{ 20 } =\frac { 360+30 }{ 20 } =\frac { 390 }{ 20 } =19.5\)

**Question 12.**

**The mean weight of 6 boys in a group is 48 kg.The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. Find the weight of sixth boy.**

**Solution:**

Mean weight of 6 boys = 48 kg

Their total weight = 48 x 6 = 288 kg

Weights of 5 boys among them, are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg

Sum of weight of 5 boys = (51 + 45 + 49 + 46 + 44) kg = 235 kg

Weight of 6th boy = 288 kg – 235 kg = 53 kg Ans.

**Question 13.**

**The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. Find the correct mean.**

**Solution:**

Mean of 50 students = 39

Total score = 39 x 50 = 1950

Now correct sum of scores = 1950 – Wrong item + Correct item= 1950 – 23 + 43

= 1950 + 20 = 1970

Correct mean = \(\frac { 1970 }{ 50 } \) = 39.4 Ans.

**Question 14.**

**The mean of 100 items was found to be 64. Later on it was discover that two items were misread as 26 and 9 instead of 36 and 90 respectively. Find the correct mean.**

**Solution:**

Mean of 100 items = 64

The sum of 100 items = 64 x 100 = 6400

New sum of 100 items = 6400 + 36 + 90 – 26 – 9 = 6526 – 35 = 6491,

Correct mean = \(\frac { 6491 }{ 100 }\) =64.91 = 64.91 Ans.

**Question 15.**

**The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number**

**Solution:**

Mean of 6 numbers = 23

Sum of 6 numbers = 23 x 6 = 138

Excluding one number, the mean of remaining 5 numbers = 20

Total of 5 numbers = 20 x 5 = 100

Excluded number = 138 – 100 = 38 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D are helpful to complete your math homework.

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