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

mean (to 2 decimals)
median (to 1 decimal)
mode (to 1 decimal)
b. range and standard deviation

range (to 1 decimal)
standard deviation (to 4 decimals)
c. should the response time of 8.3 minutes be considered an outlier in comparison to the other response times?

this observation - select your answer -doesdoes notitem 6 appear to be an outlier.

d. do the response times indicate that the city is meeting the national standards?
- select your answer -yesnoitem 7

should the city consider making changes to its response strategies?
- select your answer -yesnoitem 8

would adding more stations to areas in the city be a practical solution?
- select your answer -yesno

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).