Respuesta :
Answer:
The value is [tex]P( X > 0.50) = 0.089264[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 500
The population proportion is p = 0.47
Generally given that the sample size is sufficiently large , the mean of this sampling distribution is mathematically represented as
[tex]\mu_x = p = 0.47[/tex]
Generally the standard deviation of the sampling distribution is mathematically represented as
[tex]\sigma = \sqrt{\frac{p(1- p )}{ n} }[/tex]
=> [tex]\sigma = \sqrt{\frac{ 0.47 (1-0.47 )}{ 500 } }[/tex]
=> [tex]\sigma = 0.0223[/tex]
Gnerally the probability that the proportion of voters in the sample of Region A who support the ballot measure is greater than 0.50 is mathematically represented as
[tex]P( X > 0.50) = P( \frac{ X - \mu }{ \sigma } > \frac{ 0.50 - 0.47 }{ 0.0223 } )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P( X > 0.50) = P( Z> 1.3453 )[/tex]
From the z table the area under the normal curve to the left corresponding to 1.3453 is
[tex]P( Z> 1.3453 ) = 0.089264[/tex]
So
[tex]P( X > 0.50) = 0.089264[/tex]
Using the normal distribution and the central limit theorem, it is found that there is a 0.0901 = 9.01% probability that the proportion of voters in the sample of Region A who support the ballot measure is greater than 0.50.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
By the Central Limit Theorem, the sampling distribution of sample proportions of size n of a proportion p has [tex]\mu = p, s = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem:
- Sample of 500 voters, hence [tex]n = 500[/tex].
- The proportion is of 0.47, hence [tex]p = 0.47[/tex]
The mean and the standard deviation are:
[tex]\mu = p = 0.47[/tex]
[tex]\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.47(0.53)}{500}} = 0.0223[/tex]
The probability that the proportion of voters in the sample of Region A who support the ballot measure is greater than 0.50 is 1 subtracted by the p-value of Z when X = 0.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.5 - 0.47}{0.0223}[/tex]
[tex]Z = 1.34[/tex]
[tex]Z = 1.34[/tex] has a p-value of 0.9099.
1 - 0.9099 = 0.0901
0.0901 = 9.01% probability that the proportion of voters in the sample of Region A who support the ballot measure is greater than 0.50.
A similar problem is given at https://brainly.com/question/24663213
