The fill amount in 2-liter soft drink bottles is normally distributed with a mean of 2.0 liters and a mean of 0.05 litters. if bottles contain less than 95% of the listed net content (i. e., less than 1.90 litters) the manufacturer may be subject to penalty by the state office of consumer affairs. bottles that have a net content above 2.10 liters may cause excessive spillage upon opening. what is the probability that a randomly selected bottle will contain:
a. between 1.90 and 2.00 litters?
b. between 1.90 and 2.10 litters?
c. below 1.90 liters or above 2.10 litters?
about least how much soft drink (in litters) is contained in 99% of the bottles?
e. 99% of the bottles contain an amount that is between which two values, symmetric about the average?

a) 0.4722

b) 0.9545

c) 0.0455

d) 1.8837

e) between 1.8712 and 2.1288

Step-by-step explanation:

a) Z = (x – mean)/standard deviation

Le x be amount in liters in a bottle

P (1.9<x<2) = P((1.9– 2)/0.05 < z < (2– 2)/0.05)

P (1.9<x<2) = P( -2 < z < 0)

Then we use the z table to find the area under the curve.

P (1.9<x<2) = 0.4772

b)P (1.9<x<2.1) = P((1.9– 2)/0.05 < z < (2.1– 2)/0.05)

P (1.9<x<2.1) = P( -2 < z < 2)

Then we use the z table to find the area under the curve.

P (1.9<x<2.1) = 0.9545

c)There is a probability of 0.9545 that the content of a bottle is between 1.90 and 2.10 litters. Then the probability that the content is below 1.90 liters or above 2.10 litters is 1 – 0.9545 = 0.0455

d)z = (x– mean)/ standard deviation

x = z * standard deviation + mean

Now, z value for above the 99% has to be found using a z table.

In this case z = -2.326

x = -2.326 * .05 + 2

x = 1.8837

e)99% of the bottles contain an amount that is between which two values, symmetric about the average means a 49.5% on both sides. Z values area -2.576 and 2.576.

Replacing in the formula

x = z * standard deviation + mean

x = -2.576 * .05 + 2

x = 1.8712

x = z * standard deviation + mean

x = 2.576 * .05 + 2

x = 2.1288

99% of the bottles contain an amount that is between 1.8712 and 2.1288

( ͡° ͜ʖ ͡°)

step-by-step explanation:

just copy and paste it.

visit this site for a tutorial ʖ-tutorial-flash-cards/

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).