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1.08Discussion Details Oroumtnat Obama received 40% of the white ma vate; Ifthis I5 tnue what i5 the probability that random sample of 200 white male voters had les...

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1.08Discussion Details Oroumtnat Obama received 40% of the white ma vate; Ifthis I5 tnue what i5 the probability that random sample of 200 white male voters had less than 35% voting for Obama?Drg the orange Mag left or right t0 change the Z-score. Alternatively. enter a Z-score into the textbox and click cnicrL-score: QOOZ-scoreThe area the left of the Z value is: 0.1587The area t0 the right of the Z value iS: 0.8413Camlemthe Open Leatnirig IrrliatiYe and licensed under CC Dl-ReplyPreviousNextDi

1.08 Discussion Details Oroum tnat Obama received 40% of the white ma vate; Ifthis I5 tnue what i5 the probability that random sample of 200 white male voters had less than 35% voting for Obama? Drg the orange Mag left or right t0 change the Z-score. Alternatively. enter a Z-score into the textbox and click cnicr L-score: QOO Z-score The area the left of the Z value is: 0.1587 The area t0 the right of the Z value iS: 0.8413 Camlem the Open Leatnirig IrrliatiYe and licensed under CC Dl- Reply Previous Next Diareard Calcndar Aod ? Notmicationt



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In the 1992 presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for President Clinton. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district.
a. State the approximate distribution of X.
b. Is 1,956.8 a population mean or a sample mean? How do you know?
c. Find the probability that a randomly selected district had fewer than 1,600 votes for President Clinton. Sketch the graph and write the probability statement.
d. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton.
e. Find the third quartile for votes for President Clinton.

Okay. So for this problem we're going to apply binomial distribution just because it's given that the percentage is .5 or 50%. Um exactly five voting for Obama would be p. X. is equal to five. And then we have this beautiful little thing here where we do 10 choose five Times .5 to the power of five, 5 to the power of five. And this comes out to be equal to about 0.25. It's about a quarter. Yeah. Now for part B eight or more voting for Obama would be P X is greater than or equal to eight. Which is going to be um a sum of a couple for different things that's gonna be P is PX equals eight Plus. The probability that x equals nine Plus the probability of x equals 10 and the sample. Okay so this one is gonna be I'm gonna do this one in red. So the only numbers we're gonna be changing here It's going to be this is eight And this is gonna be 8 to This one in green. This is gonna be nine nine one And then in black p. x. equals 10. This guy is gonna be 10 10 and zero adding up all these numbers. What I end up getting is About 5.4%. And for part C were asked even if this is true, why are the probabilities that we got approximately correct? Um this we just have to do with the sample size of 10 Indiana and not representing the whole state of Indiana. We also don't really know what kind of spread this data has.

Yeah, Random variable Z has a standard normal distribution. We want to find the probability that C. is less than or equal to negative 2.15 and shade the area in on the standard normal distribution bath. To solve for this probability, we're going to rely on the Z score table which matches the scores, specifically negative Z scores for standard normal distribution to respective probabilities and therefore areas in this particular case, we're going to search for the probability that Z is less than or equal to negative 2.15 and we find that the solution to this, the probability that Z is less than equal to negative 2.15 is simply 0.158 next to sketch the area. We're going to start by remembering that the mean is equal to zero and the standard deviation is equal to one. And at the standard normal distribution typically looks like this to sketch this in, we're going to draw a vertical line at Z equals negative 2.15 far right, or rather far left of the mean, And then we're going to shape everything to the left of it. The graphics like this, note how small the area is corresponding to the small solution. We derive 0.158

Boy random variable Z has a standard normal distribution. We want to find the probability that Z is greater than or equal to negative 1.50 and shade the area and on the graph in the standard normal to start, we're going to solve for this probability by making use of what's known as a Z score table. These tables can be referenced and matt Z scores directly on two areas and probabilities for the standard normal distribution. Even though the score tables are only used for negative Z scores, we can still make use of them, relying on the symmetry of the standard normal distribution. So chooses the score table. We risk, we express this probability as one minus the probability that Z is less than or equal to negative 1.5. And plugging in for the Z score. And substituting in here, we get 0.933 to the area for probability Z is equal to negative 1.5 next to sketch this will make note of the mean and standard deviation for the standard normal, as well as the image of the graph. To turn this graph in the correct solution, we need to shade in the area that is to the right of a vertical line at Z equals negative 1.5, as is shown here with the shading and yellow.

Random variable Z has a standard normal distribution. We want to find the probability that Z falls between zero and 0.54 and showed that area in on a standard normal distribution bath. First to find this probability, we're going to make use of what's known as a Z score table, the source. These four tables can be reference and that negative Z scores directly onto the area or probability to the left of that Z score on a standard normal distribution. Well, zero and 0.54 specifically 0.54 are not negative Z scores. We rely on the symmetry of the standard normal distribution to still be able to use these four table appropriately. To do so, we re expresses probability as the probability that Z is zero minus the probability that Z is equal to negative 00.54 And using the table to look at these two scores were obtained solution of 0.2054 Now, to map this onto a graph, we start by noting the mean and standard deviation of a standard normal distribution, as well as the appearance of the graph. Here. In order to construct the correct solution graph, we draw a vertical line that Z equals zero and vehicles 00.54 and shade the area in between, as we have done here in yellow.


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