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Vsevolod [243]
2 years ago
12

At Mountain High School, the students were surveyed about their participation in band (B) and track (T). The results of the surv

ey are shown in the Venn diagram. Given that a randomly chosen student participates in band, what is the probability that the student also participates in track?
Mathematics
2 answers:
mixer [17]2 years ago
7 0
The Venn diagram is not included, so I am going to explain the procedure and you will be able to calculate the result just plugging the numbers.

From Vend diagram you will be able to conclude two numbers:

1) the total number os students surveyed that participate in band (B)

2) the total number of students surveyed that participate in track (T)

3) the total number of students surveyed that participate in both band (B) and track (T) = B∩T

4) ) the total number of students surveyed, which is: total number of students in band B + total number of students in track - less the number of students that participate in both. = B + T - (B∩T)

So, the probability that one student that participate in band (B) also participate in track (T) =

Number of students that participate in both / number of students that participate in band = (B∩T) / B

Now, you just have to find B∩T and B from the diagram and plugg in into the formula (B∩T)/B. That's it.

borishaifa [10]2 years ago
4 0
E2020 Answer is : B (9/33)
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Define z_alpha to be a z-score with an area of alpha to the right. For Example: z_0.10 means P(Z > z_0.10) = 0.10. We would a
Reptile [31]

Answer:

a) P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :z_{crit}= \pm 1.96

b) P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates \alpha/2 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

c) For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got z_{\alpha/2}=1.64

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Part a

P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :z_{crit}= \pm 1.96

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates \alpha/2 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got z_{\alpha/2}=1.64

6 0
3 years ago
J- 7.5 = 16.937 ????
Leviafan [203]
Add 16.937 and 7.5. It's basically doing the problem backwards.
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