Just before a referendum on a school budget, a local newspaper polls 402 voters to predict whether the budget will pass. Suppose the budget has the support of 52% of the voters. What is the probability that the newspaper's sample will lead it to predict defeat?

Assume that proportion of support follows normal distribution.For resulting in defeat the proportion of support should less than 0.5.Given that the proportion of support is p= 0.52.

Sample size(n)= 402

We know that standard deviation is given as

[tex]\sigma = \sqrt {\dfrac {p(1-p)}{n}}[/tex]

Now by putting the values

[tex]\sigma = \sqrt {\dfrac{0.52\times 0.48}{402}}[/tex]

σ=0.0249

Proportion of support follows normal distribution with mean is 0.52 and standard deviation is 0.0249

We know that [tex]Z=\dfrac{\bar{X}-\mu }{\sigma }[/tex]

So

[tex]Z=\dfrac{0.5-0.52 }{0.0249}[/tex]

Z= -0.80

So P(Z<-.80) =0.2118 from standard table.

So the probability of news paper to predict defeat is 21.21%

Just before a referendum on a school​ budget, a local newspaper polls 402 voters to predict whether the budget will pass. Suppose the budget has the support of 52​% of the voters. What is the probability that the​ newspaper's sample will lead it to predict​ defeat?

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Just before a referendum on a school budget, a local newspaper polls 402 voters to predict whether the budget will pass. Suppose the budget has the support of 52% of the voters. What is the probability that the newspaper's sample will lead it to predict defeat?