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2 years ago

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From 2003 onward, the number of daily visitors to a website increased by 200% every two years. So, for example, the number of vi

sitors in 2011 was 200% more than the number of visitors in 2009.
In what year was the number of daily visitors $800\%$ more than the number of daily visitors in 2003?

1 answer:

SSSSS [86.1K]2 years ago

4 0

It takes 4 years for number of daily visitors to be 800% more than the number of daily visitors in 2003

<em><u>Solution:</u></em>

Let the number of visitors in 2003 be 100

From 2003 onward, the number of daily visitors to a website increased by 200% every two years

<em><u>Therefore, Number of visitors in 2005 is given as:</u></em>

Visitors in 2005 = 100 + 200 % of 100

$Visitors\ in\ 2005 = 100 + \frac{200}{100} \times 100\\$

Visitors in 2005 is 300

<em><u>Similarly, in 2007, the number of visitors is given as:</u></em>

Visitors in 2007 = 300 + 200 % of 300

Visitors in 2007 = 300 + 600

Visitors in 2007 = 900

<em><u>Now find the percentgae increase from 2003 to 2007</u></em>

$\text { Percentage increase }=\frac{\text {visitors in } 2007-\text {visitors in } 2003}{\text {visitors in } 2003} \times 100$

$\text{percentgae increase} = \frac{900 - 100}{100} \times 100\\\\\text{percentgae increase} = 800$

Thus it is a 800 % increase

Number of years = 2007 - 2003 = 4 years

Thus it takes 4 years for number of daily visitors to be 800% more than the number of daily visitors in 2003

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And we can find the usual limits with:

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And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

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Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

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Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

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And if we replace we got:

$s= 0.948$

Part c

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We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

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Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

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