Respuesta :
Answer:
Probability that the proportion of politicians who are lawyers will be less than 51% is 0.0022.
Step-by-step explanation:
We are given that 57% of politicians are lawyers. A random sample of size 564 is selected.
Let [tex]\hat p[/tex] = sample proportion of politicians who are lawyers
The z-score probability distribution for sample 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] = sample proportion
p = population proportion = 57%
n = sample size = 564
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.
So, probability that the proportion of politicians who are lawyers will be less than 51% is given by = P( [tex]\hat p[/tex] < 0.51)
P( [tex]\hat p[/tex] < 0.51) = P( [tex]\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]\frac{ 0.51-0.57}{\sqrt{\frac{0.51(1-0.51)}{564} } }[/tex] ) = P(Z < -2.85) = 1 - P(Z [tex]\leq[/tex] 2.85)
= 1 - 0.9978 = 0.0022
Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.85 in the z table which has an area of 0.9978.
Therefore, probability that the proportion of politicians who are lawyers will be less than 51% is 0.0022.
