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3 years ago

In a simple random sample of 1300 voters, 19% said they would prefer to vote online.

Mathematics

1 answer:

dsp733 years ago

8 0

Solution:

Formula for calculation of 99% confidence interval is given by:

= $P \pm (\text{Z value} )\sqrt\frac {P\times(1-P)}{N}$

P= Sample Proportion

N =Sample Size

Also,Number of voters who vote online= 19 % of 1300=19 × 13=247

Sample Population= $=\frac{247}{1300}$ =0.19

Z value for 99% confidence interval=2.58

The number of voters who vote online, who are in the 99% confidence interval for the percent of all voters who would prefer to vote online= $0.19 \pm (2.58)\sqrt\frac{0.19 \times (1-0.19)}{1300}\\\\ =0.19 + (2.58)\sqrt\frac{0.19 \times (1-0.19)}{1300} {\text{or}} 0.19 - (2.58)\sqrt\frac{0.19 \times (1-0.19)}{1300}\\\\ 0.2180 {\text{or}} 0.1619$

So, Percent of voters , who are in the 99% confidence interval for the percent of all voters who would prefer to vote online= 100 × 0.2180 or 100 × 0.1619=21.80 % or 16.19 %

That is between, 16.19 % and 21.80 %.

→→Option (A) 19.8 % and Option (B)17.4 % are in 99 % confidence intervals.

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riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

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