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galina1969 [7]
3 years ago
8

A group of 10 students received the following marks for a test: 58, 89,65, 78, 55,26,93,46,43,59. The standard deviation of thei

r marks is (to two decimal places)
Mathematics
1 answer:
Anna35 [415]3 years ago
6 0

Answer:19.84

Step-by-step explanation:

S. D = √(Variance)

To find the standard deviation, we need to first find the variance

variance = ∑(x-x⁻)² / n

where x⁻ = mean

n=no. of data = 10

∑ means summation

again, we to find the mean

mean(x⁻) = 58+89+65+78+55+26+93+46+43+

59 /10

=612/10

=61.2

mean(x⁻) =61.2

x x⁻ (x-x⁻)²

58 61.2 10.24

89 61.2 772.84

65 61.2 14.44

78 61.2 282.24

55 61.2 38.44

26 61.2 1239.04

93 61.2 1011.24

46 61.2 231.04

43 61.2 331.24

59 61.2 4.84

∑(x-x⁻)² = 10.24+772.84+14.44+282.24+38.44+1239.04+1011.24+231.04+331.25+ 4.84 = 3935.6

variance = ∑(x-x⁻)² / n

=3935.6 / 10

=393.56

variance = 393.56

S. D = √(Variance)

=√(393.56)

=19.8383

S.D = 19.84 to two decimal place

Therefore the standard deviation is 19. 84

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A worker was paid a salary of $10,500 in 1985. Each year, a salary increase of 6% of the previous year's salary was awarded. How
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In the year 1985 the worker earned <span>$10,500. 

</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

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$10,500(1+0.06) [1+0.06]=$10,500(1.06)^2.


Similarly we can check that in the year 1987 the worker earned $10,500(1.06)^3, which makes the pattern clear. 


We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

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Factorizing, we have

=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]

We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

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The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating (1.06)^{26}.


Answer: D



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