How much does a sleeping bag cost? let's say you want a sleeping bag that should keep you warm in temperatures from 20°f to 45°f. a random sample of prices ($) for sleeping bags in this temperature range is given below. assume that the population of x values has an approximately normal distribution. 35 85 105 40 100 50 30 23 100 110 105 95 105 60 110 120 95 90 60 70 (a) use a calculator with mean and sample standard deviation keys to find the sample mean price x and sample standard deviation s. (round your answers to two decimal places.) x = $ 79.4 correct: your answer is correct. s = $ 30.62 correct: your answer is correct. (b) using the given data as representative of the population of prices of all summer sleeping bags, find a 90% confidence interval for the mean price μ of all summer sleeping bags. (round your answers to two decimal places.)

a) Sample Mean = 79.4

Sample standard deviation = 30.62

b) 90% Confidence interval:  (67.56 ,91.24)

Step-by-step explanation:

We are given the following in the question:

Prices for sleeping bags has an approximately normal distribution.

We are given the following sample:

35, 85, 105, 40, 100, 50, 30, 23, 100, 110, 105, 95, 105, 60, 110, 120, 95, 90, 60, 70

a)

Formula:

where are data points, is the mean and n is the number of observations.  

Sum of squares of differences = 1971.36 + 31.36 + 655.36 + 1552.36 + 424.36 + 864.36 + 2440.36 + 3180.96 + 424.36 + 936.36 + 655.36 + 243.36 + 655.36 + 376.36 + 936.36 + 1648.36 + 243.36 + 112.36 + 376.36 + 88.36 = 17816.8

b) 90% Confidence interval:  

Putting the values, we get,  

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

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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).