Respuesta :
Answer:
0.0465 = 4.65% probability that the proportion of persons with a college degree will be less than 44%
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 780, p = 0.47[/tex]
So
[tex]\mu = E(X) = np = 780*0.47 = 366.6[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{780*0.47*0.53} = 13.94[/tex]
What is the probability that the proportion of persons with a college degree will be less than 44%?
This is the pvalue of Z when X = 0.44*780 = 343.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{343.2 - 366.6}{13.94}[/tex]
[tex]Z = -1.68[/tex]
[tex]Z = -1.68[/tex] has a pvalue of 0.0465
0.0465 = 4.65% probability that the proportion of persons with a college degree will be less than 44%
Answer:
The probability that the proportion of persons with a college degree will be less than 44% is 0.0455.
Step-by-step explanation:
We are given that 47% of the population has a college degree and a random sample of size 780 is selected.
Let p = % of population who has a college degree = 47%
The z-score probability distribution for proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = proportion of persons with a college degree in a sample of 780 selected
n = sample size = 780
So, probability that the proportion of persons with a college degree will be less than 44% is given by = P([tex]\hat p[/tex] < 0.44)
P([tex]\hat p[/tex] < 0.44) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]\frac{0.44-0.47}{\sqrt{\frac{0.44(1-0.44)}{780} } }[/tex] ) = P(Z < -1.69) = 1 - P(Z [tex]\leq[/tex] 1.69)
= 1 - 0.95449 = 0.0455
The above probability is calculated using z table by looking at value of x = 1.69 in the z table which have an area of 0.95449.
Therefore, probability that the proportion of persons with a college degree will be less than 44% is 0.0455.