A sequence of 50 numerical measurements x_1, x_2, \dots, x_{50} is collected. Each measure… — GRE General Test

A sequence of 50 numerical measurements $x_1, x_2, \dots, x_{50}$ is collected. Each measurement $x_i$ is rounded to the nearest tenth to produce a rounded value $r_i$ . The sum of the 50 rounded values, $\sum_{i=1}^{50} r_i$ , is equal to $250.0$ . If $S$ represents the true sum of the unrounded measurements $\sum_{i=1}^{50} x_i$ , what is the maximum possible percent error of the rounded sum relative to the true sum $S$ , rounded to the nearest hundredth of a percent?

A
$0.99\%$
B
$1.00\%$
$1.01\%$ Cevap
D
$2.00\%$
E
$2.04\%$

Cevap

The maximum possible percent error of the rounded sum relative to the true sum is

1.01\%

When rounding numbers to the nearest tenth, the maximum error for each number is

0.05

. For 50 numbers, the maximum possible error in the sum is

50 \times 0.05 = 2.5

. The true sum

S

therefore lies in the range

[247.5, 252.5]

. To maximize the percent error relative to

S

, defined as

\frac{|250.0 - S|}{S} \times 100\%

, we use the maximum numerator

2.5

and the smallest possible denominator

S = 247.5

. This yields

\frac{2.5}{247.5} \times 100\% \approx 1.01\%

Adım Adım Çözüm

Determine the maximum rounding error for a single term

Maximum error per measurement is

|x_i - r_i| \le 0.05

When rounding to the nearest tenth, any value within

0.05

of the rounded value rounds to that tenth.

Calculate the maximum cumulative error for the sequence sum

Maximum total error

= 50 \times 0.05 = 2.5

The maximum difference between the true sum

S

and the rounded sum

250.0

occurs when all individual rounding errors accumulate in the same direction.

Find the range of possible true sum values

S

247.5 \le S \le 252.5

Subtracting and adding the maximum error of

2.5

from the rounded sum

250.0

establishes the bounds for

S

Set up and maximize the percent error expression

Max percent error occurs at minimum

S = 247.5

, giving

\frac{2.5}{247.5} \times 100\% \approx 1.0101\%

Percent error relative to

S

is given by

\frac{|250.0 - S|}{S} \times 100\%

. To maximize this ratio, we divide the maximum numerator

2.5

by the smallest positive denominator

S = 247.5

Anahtar Kavram

Error propagation in sequence sums and optimizing percent error base values

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

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