Respuesta :
Answer:
a) X = N(1955,556)
b) 1955 is the population mean
c) P(X <1911) = 0.4685
d) P(1868 < X <2105) = 0.1685
e) 95th percentile = 2870 (nearest whole number)
Step-by-step explanation:
a) Average number of votes per district, μ = 1955
The standard deviation, [tex]\sigma = 556[/tex]
The distribution of X will take the form [tex]N(\mu, \sigma)[/tex]
Therefore the distribution of X is N(1955,556)
b)1955 is the average of all the votes president Clinton had per district(i.e. the mean of all the values of the population) and not the mean of collected samples.
1955 is the population mean
c) Probability that a randomly selected district had fewer than 1911 votes for president Clinton
[tex]P(X < 1911) = P ( z < \frac{x - \mu}{\sigma} )[/tex]
z=(1911-1955)/556
z=-0.079
From the probability distribution table
P(z<-0.079)=0.4685
Therefore, P(X <1911) = 0.4685
d)probability that a randomly selected district had between 1868 and 2105 votes for President Clinton
[tex]P(x_{1} < X <x_{2}) = P(z_{2} < \frac{x_{2} - \mu }{\sigma} )- P(z_{1} < \frac{x_{1} - \mu }{\sigma} )[/tex]
[tex]P(1868 < X <2105) = P(z_{2} < \frac{2105 - 1955 }{556} )- P(z_{1} < \frac{1868 - 1955 }{556} )\\P(1868 < X <2105) = P(z_{2} < 0.27)- P(z_{1} < -0.16 )[/tex]
P(1868 < X <2105) = 0.6063 - 0.4378
P(1868 < X <2105) = 0.1685
e) Find the 95th percentile for votes for President Clinton
[tex]P_{95} = \mu + 1.645 \sigma\\P_{95} = 1955 + 1.645(556)\\P_{95} = 2869.62\\P_{95} = 2970[/tex]
P95=1955+1.645*556=2870
