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ArbitrLikvidat [17]
2 years ago
6

The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi

dents of a certain county included 1210 who were over the age of 25, and 1030 of them were high school graduates.
(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

Mean =

Standard deviation =

(b) Is that county result of 1030 unusually high, or low, or neither? (Enter HIGH or LOW or NEITHER)
Mathematics
1 answer:
SVETLANKA909090 [29]2 years ago
4 0

Answer:

a) Mean = 1030; Standard deviation = 12.38.

b) The county result is unusually high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

This first question is a binomial propability distribution.

We have a sample of 1210 Amricans, so n = 1210.

The mean of the sample is 1030.

The probability of a success is \pi = \frac{1030}{1210} = 0.8512.

The standard deviation of the sample is s = \sqrt{n\pi(1-\pi)} = \sqrt{1210*0.8512*0.1488} = 12.38

(b) Is that county result of 1030 unusually high, or low, or neither?

The first step is find the zscore when X = 1030.

Then we find the pvalue of this zscore.

If this pvalue is bigger than 0.95, the county result is unusually high.

If this pvalue is smaller than 0.05, the county result is unusually low.

Otherwise, it is neither.

The national mean is 82%. So,

\mu = 0.82(1210) = 992.2

Z = \frac{X - \mu}{\sigma}

Z = \frac{1030 - 992.2}{12.38}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989.This means that the county result is unusually high.

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Answer:

a. 0.6

b. not independent

c. 0.1

d. 0.4

e. 0.3

Step-by-step explanation:

a.

P(passing first course)=P(C1)=0.7

P(passing second course)=P(C2)=0.8

P(passing at least one course)=P(C1∪C2)=0.9

P( passes both courses)=P(C1∩C2)=?

We know that

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A∩B)=P(A)+P(B)-P(A∪B)

So,

P( passes both courses)=P(C1∩C2)=P(C1)+P(C2)-P(C1∪C2)

P( passes both courses)=P(C1∩C2)=0.7+0.8-0.9

P( passes both courses)=P(C1∩C2)=0.6

Thus, the probability she passes both courses is 0.6.

b.

The event of passing one course is independent of passing another course if

P(C1∩C2)=P(C1)*P(C2)

P(C1)*P(C2)=0.7*0.8=0.56

P(C1∩C2)=0.6

As,

0.6≠0.56

P(C1∩C2)≠P(C1)*P(C2),

So, the event of passing one course is dependent of passing another course.

c.

P(not passing either course)=P(C1∪C2)'=1-P(C1∪C2)

P(not passing either course)=P(C1∪C2)'=1-0.9

P(not passing either course)=P(C1∪C2)'=0.1

Thus, the probability of not passing either course is 0.1.

d.

P(not passing both courses)=P(C1∩C2)'=1-P(C1∩C2)

P(not passing both courses)=P(C1∩C2)'=1-0.6

P(not passing both courses)=P(C1∩C2)'=0.4

Thus, the probability of not passing both courses is 0.4.

e.

P(passing exactly one course)=?

P(passing exactly course 1)=P(C1)-P(C1∩C2)=0.7-0.6=0.1

P(passing exactly course 2)=P(C2)-P(C1∩C2)=0.8-0.6=0.2

P(passing exactly one course)=P(passing exactly course 1)+P(passing exactly course 2)

P(passing exactly one course)=0.1+0.2

P(passing exactly one course)=0.3

Thus, the probability of passing exactly one course is 0.3.

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A rocket leaves the surface of Earth at time t=0 and travels straight up from the surface. The height, in feet, of the rocket ab
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Answer:

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Where b is the next value of t0, and a is the previous value of t0. This is because t0 is the middle point in this segment.

Then:

if t0 = 100s

AS(100s) = (400ft - 0ft)/(200s - 0s) =   2ft/s

if t0 = 200s

AS(200s) = (1360ft - 50ft)/(300s - 100s) = 6.55 ft/s

if t0 = 300s

AS(300s) = (3200ft - 400ft)/(400s - 200s) =  14ft/s

if t0 = 400s

AS(400s) = (6250s - 1360s)/(500s - 300s) = 24.45 ft/s

So for the given options, t = 400s is the one where the velocity seems to be the biggest.

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Then we can conclude that the rocket is accelerating upwards, then as larger is the value of t, bigger will be the average velocity at that point.

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Answer:

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With traffic :

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