According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for medical aid varies dramatically across Los Angeles (LA Times website, November 2012). Under national standards adopted by the Los Angeles Fire Department, rescuers are supposed to arrive within six minutes to almost all medical emergencies. But the Times analysis found that in affluent hillside communities stretching from Griffith Park to Pacific Palisades, firefighters failed to hit that mark nearly 85% of the time. The following data show the response times, in minutes, for 10 emergency calls in the Griffith Park neighborhood.

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2
Based on this sample of ten response times, compute the following descriptive statistics:

a. Mean, median, and mode
b. Range and standard deviation
c. Should the response time of 8.3 minutes be considered an outlier in comparison to the other response times?
d. Do the response times indicate that the city is meeting the national standards?Should the city consider making changes to its response strategies?Would adding more stations to areas in the city be a practical solution?

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:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And if we replace we got: [tex]\bar X=10.65[/tex]

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:

[tex] Median =\frac{10.7+10.7}{2}=10.7[/tex]

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

[tex] Mode= 10.7[/tex]

Part b

The range is defined as [tex] Range =Max-Min[/tex] and if we replace we got:

[tex] Range = 11.8-8.3=3.5[/tex]

We can calculate the standard deviation with the following formula:

[tex] s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

And if we replace we got:

[tex] s= 0.948[/tex]

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

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

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got [tex] Q_3 = \frac{10.9+11.2}{2}=11.05[/tex]

Then we can find the IQR like this:

[tex] IQR = Q_3 -Q_1 = 11.05-10.55=0.5[/tex]

And we can find the usual limits with:

[tex] Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8[/tex]

[tex] Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3[/tex]

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

Part 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.

astha8579 astha8579 · Answer 2 · 2022-01-10T06:38:15+01:00

Given 10 responses are:

11.8, 10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

Calculations of descriptive statistics can be done as follows:

Part a.

Mean, Median and mode:

Sorting the data in ascending order:

8.3, 10.3, 10.5, 10.6, 10.7, 10.7, 10.9, 11.2, 11.5, 11.8

Thus, since 10.7 occurs maximum number of times, its the mode.

The middle pair of values are 10.7 and 10.7, thus median is also 10.7

The mean of given data is calculated as:

[tex]\overline{x} = \dfrac{(8.3+ 10.3+ 10.5+ 10.6+10.7+ 10.7+ 10.9+11.2+11.5+11.8)}{10}\\\overline{x} = \dfrac{106.5}{10}\\\overline{x} = 10.65[/tex]

Thus, mean is 10.65

Part b.

Range and standard deviation:

The range is given as maximum - minimum observation, thus it being 11.8-8.3 = 3.5

The standard deviation is calculated as:

[tex]\sigma = \sqrt{\sum{\dfrac{(x_i - \overline{x})^2}{n}}}\\\sigma = \sqrt{\dfrac{8.085}{10}}\\\sigma = \sqrt{ 0.8085}\\\sigma = 0.899..[/tex]

Thus, standard deviation for given data is 0.899

Part c.

[tex]IQR= Q_3 - Q_1 = 11.05 - 10.55 = 0.5[/tex]

And [tex]\rm lower\: limit = 10.5 - 1.5 \times IQR = 10.5 - 1.5 \times 0.5 = 9.8[/tex]

And since 8.3 < 9.8, thus its lower than lower limit itself, thus being an outlier in comparison to other responses.

Part d.

The data lies around 10.65 and has maximum of 3.5 as range. Since its in minutes and the standard response time is around 9 minutes, thus its meeting the national standards.

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