What percentage of the general U. S. population have bachelor's degrees? The Statistical Abstract of the United States, 120th Edition, gives the percentage of bachelor's degrees by state. For convenience, the data are sorted in increasing order. 17 18 18 18 19 20 20 20 21 21 21 21 22 22 22 22 22 22 23 23 24 24 24 24 24 24 24 24 25 26 26 26 26 26 26 27 27 27 27 27 28 28 29 31 31 32 32 34 35 38

(a) Select the box-and-whisker plot.
(b) Find the interquartile range.
IQR =
(c) Illinois has a bachelor's degree percentage rate of about 26%. Into what quarter does this rate fall?

below Q1between Q1 and the median between the median and Q3above Q3

a) Figure attached

b)

c) between the median and Q3

Step-by-step explanation:

For this case we have the following data:

17 18 18 18 19 20 20 20 21 21 21 21 22 22 22 22 22 22 23 23 24 24 24 24 24 24 24 24 25 26 26 26 26 26 26 27 27 27 27 27 28 28 29 31 31 32 32 34 35 38

Part a

We can construct the boxplot with the following R code:

> x<-c(17, 18, 18, 18, 19, 20, 20, 20, 21 ,21, 21, 21, 22, 22, 22, 22, 22, 22, 23, 23, 24, 24, 24, 24, 24, 24, 24, 24, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 28, 28, 29, 31, 31, 32, 32, 34, 35, 38)

> boxplot(x,main="Boxplot of data")

> summary(x)

  Min. 1st Qu.  Median    Mean 3rd Qu.    Max.  

 17.00   22.00   24.00   24.76   27.00   38.00

The results are on the figure attached.

The mean >Median so then we can conclude that the distribution is slightly skewed to the right.

Part b

For this case since we have 50 data values in order to find the the Q1 we use the following values 17, 18, 18, 18, 19, 20, 20, 20, 21 ,21, 21, 21, 22, 22, 22, 22, 22, 22, 23, 23, 24, 24, 24, 24, 24, 24

And the Q1 would be the average between position 13 and 14 and we got:

For the Q3 we need to use the following data values: 24, 24, 24, 24, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 28, 28, 29, 31, 31, 32, 32, 34, 35, 38

And the Q3 would be the average between position 13 and 14 and we got:

So then

Part c

For this case we have a value of 26 and if we see this value is between the meadian and Q3 since the median = 24 and Q3 = 27

between the median and Q3

b

step-by-step explanation:

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)

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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).