Respuesta :
Answer: 0.6368
Step-by-step explanation:
Given : The proportion of z-test for population mean : [tex]\hat{p}= 0.80[/tex]
Sample size : [tex]n=45[/tex]
We assume that this is normally distributed.
The test statistic for population proportion :-
[tex]z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}[/tex]
Now, for [tex]p_0=0.82[/tex], we have
[tex]z=\dfrac{0.80-0.82}{\sqrt{\dfrac{0.82(1-0.82)}{45}}}\\\\\approx-0.35[/tex]
By using the standard normal distribution table , the probability that at least 82% of the surveyed students will be attending college will be :-
[tex]P(p\geq0.82)=P(z\geq-0.35)\\\\=1-P(z<-0.35)\\\\=1-0.3631693\approx0.6368307\approx0.6368[/tex]
Hence, the probability that at least 82% of the surveyed students will be attending college = 0.6368
The probability that at least 82% of the surveyed students will be attending college is 0.6368 approximately.
How to find the corresponding value of z test statistic for a value of proportion random variable?
Suppose the population proportion mean be denoted by [tex]\hat{p}[/tex] and the population proportion random variable be denoted by [tex]p[/tex]
Then, the z-test statistic's value for [tex]p = p_0[/tex] Â for sample of size n is specified by
[tex]z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}[/tex]
For this case, if we take;
p = random variable taking values of the proportion of number of students graduating from high school from the sample of size 40,
then, we have:
[tex]\hat{p} = 80\% = 0.80[/tex] = mean  population proportion
and we want to know the probability for p [tex]\geq[/tex] 82% = 0.82 for sample size n = 45
The z score or z-test statistic for p = 0.82 (see its equal right now and not greater or equal) is obtained as:
[tex]z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}\\\\z = \dfrac{0.80 - 0.82}{\sqrt{\dfrac{0.82(0.18)}{45}}} \approx -0.34[/tex]
The probability for p [tex]\geq[/tex] 82% = 0.82 for sample size n = 45, is therefore can be written as:
[tex]P(Z \geq z \approx -0.35)[/tex] where Z is the standard normal variate.
This is rewritten as:
[tex]P(Z \geq -0.35) = 1 - P(Z < -0.35)[/tex]
From z-tables, the p value for Z Â = -0.35 is 0.3632, this is the probability of [tex]P(Z \leq -0.35)[/tex]
Since individual values have 0 probability in continuous distribution, therefore we get:
[tex]P(Z \geq -0.35) = 1 - P(Z < -0.35) = 1 - P(Z \leq -0.35) \approx 1 - 0.3632\\P(Z \gqe -0.35) \approx 0.6368[/tex]
Thus, the probability that at least 82% of the surveyed students will be attending college is 0.6368 approximately.
Learn more about sample proportion here:
https://brainly.com/question/16105218
