1. the mean of 7 numbers is 63. what is the sum of the numbers?

Mean = (Sum of numbers)/ Total Numbers

Here,

Mean = 63

Total Numbers = 7

63 = (Sum of numbers)/ 7

Sum of numbers = 63 x 7

Sum of numbers = 441

2. out of 100 numbers, 16 where 5’s, 21 where 6’s, 30 where 7’s and the rest where 8. find the mean of the numbers.

Mean = (Sum of numbers)/ Total Numbers

16 where 5’s = 16 x 5 = 80

21 where 6’s = 21 x 6 = 126

30 where 7’s = 30 x 7 = 210

rest where 8 , where rest = 100 – 16 – 21 – 30 = 33

so, rest where 8 = 33 x 8 = 264

Sum of numbers = 80 + 126 + 210 + 264 = 680

Mean = (Sum of numbers)/ Total Numbers

Mean = 680/100

Mean = 6.8

3. 30 students in a class average 80% on certain examination. 20 others average 90. what is the average of the students?

30 students in a class average 80%

Mean = (Sum of Percentages)/ Total Numbers

80% = (Sum of Percentages)/30

Sum of Percentages = 2400%

20 others average 90%

Mean = (Sum of Percentages)/ Total Numbers

90% = (Sum of Percentages)/20

Sum of Percentages = 1800%

So,

Mean = (Sum of Percentages)/ Total Numbers

Mean = (2400 %+ 1800%)/(30+20)

Mean = (4200%)/50

Mean = 84%

4. if the average annual income of 15 workers is 66,000 and 6 of the workers made P30,000 a year, what is the average annual income of the remaining 9 workers?

Average annual income 15 workers is 66,000

Average annual income = (Sum of incomes)/ Total Numbers

66,000 = (Sum of incomes of 15 workers)/15

Sum of incomes of 15 workers = 990000

Average annual income 6 workers is 30,000

Average annual income = (Sum of incomes)/ Total Numbers

30,000 = (Sum of incomes of 6 workers)/6

Sum of incomes of 6 workers = 180000

Sum of incomes of 9 workers = (Sum of incomes of 15 workers) – (Sum of incomes of 6 workers)

Sum of incomes of 9 workers = 990000 -180000

Sum of incomes of 9 workers = 810000

Average annual income 9 workers = (Sum of incomes of 9 workers)/ Total Numbers

Average annual income 9 workers = (Sum of incomes of 9 workers)/9

Average annual income 9 workers = 810000/9

Average annual income 9 workers = 90000

5. anna has an average of 87% on 5 examinations in statistics. what must she get in the 6th examination to average 88% on the 6 exams?

Anna has an average of 87% on 5 examinations

Mean = (Sum of Percentages)/ Total Numbers

87% = (Sum of Percentages)/ 5

Sum of Percentages = 87 x 5

Sum of Percentages = 435%

Anna wants to average 88% on the 6 exams

Consider Anna gets “x%” on the 6th exam

Sum of Percentages = 435% + x%

Mean = (Sum of Percentages)/ Total Numbers

88% = (435% + x%)/6

528% = 435% + x%

x% = 528% – 435%

x% = 93%

at the beginning you should understand that there are 6 numbers on a die. a) There are 3 even numbers (2, 4 and six) and 3 numbers extra desirable than 3 (4, 5 and six) on a die. So, altogether, we’ve 4 numbers that have compatibility this condition (2, 4, 5, and six), so the probability is 4/6 = 2/3 (Ans) b) There are 3 weird and wonderful numbers (a million, 3 and 5) and 3 numbers decrease than 4 (a million, 2 and 3) on a die. So, altogether, we’ve 4 numbers that have compatibility this condition (a million, 2, 3 and 5), so the probability is 4/6 = 2/3 (Ans) c) There are 3 weird and wonderful numbers (a million, 3 and 5) and 3 numbers extra desirable than 3 (4, 5 and six) on a die. So, altogether, we’ve 5 numbers that have compatibility this condition (a million, 3, 4, 5 and six), so the probability is 5/6 (Ans) d) There are 3 weird and wonderful numbers (a million, 3 and 5), yet there is not any quantity extra desirable than 6. So, altogether, we’ve 3 numbers that have compatibility this condition (a million, 3 and 5), so the probability is 3/6 = a million/2(Ans) e) There are 3 weird and wonderful numbers (a million, 3 and 5) and 5 numbers decrease than 6 (a million, 2, 3, 4 and 5) on a die. So, altogether, we’ve 5 numbers that have compatibility this condition (a million, 2, 3, 4 and 5), so the probability is 5/6 (Ans) f) There are 3 even numbers (2, 4 and six) and four numbers decrease than 5 (a million, 2, 3 and four) on a die. So, altogether, we’ve 5 numbers that have compatibility this condition (a million, 2, 3, 4 and six), so the probability is 5/6 (Ans)

1. sum=7*63=441

2.mean=((16*5)+(21*6)+(30*7)+(33*8))/100=6.8

3.average =48+36=84

4. 66*15-6*30=810(thousand)

810/9=90,000

5.88*6-87*5=93

1) mean=sum of no./total no.

let some of no.=x

x/7=63

x=63*7

x=441

2) mean=sum/total no.

=[16(5)+21(6)+30(7)+33(8)]/100

=[80+126+210+264]/100

=6.8