Suppose a candidate for public office is favored by only 48% of the voters. if a sample survey randomly selects 2500 voters, the percentage in the sample who favor the candidate can be thought of as a measurement from a normal curve with a mean of 48% and a standard deviation of 1%. based on this information, how often would such a survey show that 50% or more of the sample favored the candidate?

Mean=0.48
standard deviation=0.01
thus using the z-score:
P(x>0.5) we shall have the following:
z=(0.5-0.48)/0.01=2
thus
P(x>0.5)
=1-P(x<0.5)
=1-P(z<2)
=1-0.9772
=0.0228

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bhoopendrasisodiya34 bhoopendrasisodiya34 · Answer 2 · 2022-03-09T07:45:05+01:00

For the sample survey of voters 0.0228 is the oftenest that which show that 50% or more of the sample favored the candidate.

What is random sample?

Random sample is the way to choose a number or sample in such a manner that each of the sample of the group has an equal probability to be chosen.

A candidate for public office is favored by only 48% of the voters. Thus the mean value of the survey is,

[tex]m=0.48[/tex]

The sample survey randomly selects 2500 voters.

The percentage in the sample who favor the candidate can be thought of as a measurement from a normal curve with a mean of 48% and a standard deviation of 1%. Thus the standard daviation is,

[tex]\sigma =0.01[/tex]

Thus, the z score can be found out as,

[tex]z=\dfrac{0.5-m}{\sigma}\\z=\dfrac{0.5-0.48}{0.01}\\z=2[/tex]

As the 50% sample favored the candidate. Therefore, for the p value[tex]P(x > 0.5)[/tex],

[tex]P(x > 0.5)=1-P(x > 0.5)\\P(x > 0.5)=1-P(z < 2)\\P(x > 0.5)=1-0.9772\\P(x > 0.5)=0.0228[/tex]

Hence, for the sample survey of voters 0.0228 is the oftenest that which show that 50% or more of the sample favored the candidate.

Learn more about the random sample here;

https://brainly.com/question/17831271