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Suppose that 55 percent of the voters in a particular region support a candidate. find the probability that a sample of 1,000 voters would yield a sample proportion in favor of the candidate within 2 percentage points of the actual proportion.

Respuesta :

Here the original proportion is [tex] p=0.55 [/tex]. The proportion approximately follows normal distribution. Mean is [tex] \mu=p=0.55 [/tex].

Standard deviation is [tex] \sigma=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.55(1-0.55)}{1000}}=0.0157
[/tex]

The required probability is

[tex] P(0.55-0.02<X<0.55+0.02)=P(0.53<X<0.57) [/tex]

[tex] P(0.53<X<0.57)=P(\frac{0.53-0.55}{0.0157}<Z<\frac{0.57-0.55}{0.0157}) [/tex]

[tex] P(0.53<X<0.57)=P(-1.274<Z<1.274) [/tex]

Using standard normal tables,

[tex] P(0.53<X<0.57)=0.7973 [/tex]

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