## RS Aggarwal Solutions Chapter 14 Statistics Ex 14D in PDF for Free Download

Contents

**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 = = = = 4.5

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

∴ Mean = = = 10

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

∴ Mean = = = 5.6

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

∴ Mean = = = 7

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

∴ Mean = = = 20

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

∴ Mean = = = 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

**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

**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 = =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

= = 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 ****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

=> 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 =

**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 = = 40

=> x1+x2+x3+….+x12 = 40 X 12 =480

Now,new numbers are

Mean =

=

= = 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 = = 18

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

Mean of the new numbers =

=

**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 = = 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 = =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.

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