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Svetach [21]
4 years ago
13

Julia surveyed her friends to find the number of hours they spend on homework during the week. The data from her survey is displ

ayed in the first table. She then took a random sample of five responses from the population as shown in the second table. Compare the mean of the population with the mean of the given sample.
Population Data
4
5
3
1
3
2
2
3
5
7
3
6
3
0
1
5
0
4
3
6

Sample Data
5
4
6
2
1

What is the difference between the mean of the sample and the mean of the population?
0.2
0.3
0.4
0.5
Mathematics
2 answers:
Nutka1998 [239]4 years ago
7 0
The Population mean is 3.3 and the Sample mean is 3.6, 3.6-3.3 is 0.3
Marysya12 [62]4 years ago
4 0

Solution:

we are given with two kind of data set , one is Population data and the other is sample data.

we have been asked to Compare the mean of the population with the mean of the given sample.

Given Population Data
is

4 , 5 , 3 ,1 ,3 ,2,2 ,3 ,5 ,7 ,3 ,6 ,3 ,0 ,1 ,5 ,0 ,4 ,3 ,6

Its mean can be calculated as below

Mean=\frac{4+5+3+1+3+2+2+3+5+7+3+6+3+0+1+5+0+4+3+6}{20} \\\\\Mean=\frac{66}{20}=3.3\\

Now the Sample data is

5,4,6,2,1

Mean=\frac{5+4+6+2+1}{5}\\\\Mean=\frac{18}{5}=  3.6\\

Hence the value of difference of the Mean is =3.6-3.3=0.3




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Please help !!<br>[90 points if correct]​
3241004551 [841]

Answer:

2201.8348 ; 3 ; x / (1 + 0.01)

Step-by-step explanation:

1)

Final amount (A) = 2400 ; rate (r) = 6% = 0.06, time, t = 1.5 years

Sum = principal = p

Using the relation :

A = p(1 + rt)

2400 = p(1 + 0.06(1.5))

2400 = p(1 + 0.09)

2400 = p(1.09)

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p = 2201.8348

2.)

12000 amount to 15600 at 10% simple interest

A = p(1 + rt)

15600 = 12000(1 + 0.1t)

15600 = 12000 + 1200t

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t = 3600 / 1200

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3.)

A = p(1 + rt)

x = p(1 + x/100 * 1/x)

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x = p(1 + 1 / 100)

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x / (1 + 0.01)

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3 years ago
Joe used a project management software package and has determined the following results for a given project.: Expected completio
statuscvo [17]

Answer:

0.1151 = 11.51% probability of completing the project over 20 days.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Expected completion time of the project = 22 days.

Variance of project completion time = 2.77

This means that \mu = 22, \sigma = \sqrt{2.77}

What is the probability of completing the project over 20 days?

This is the p-value of Z when X = 20, so:

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

Z = \frac{20 - 22}{\sqrt{2.77}}

Z = -1.2

Z = -1.2 has a p-value of 0.1151.

0.1151 = 11.51% probability of completing the project over 20 days.

4 0
3 years ago
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