A worldwide organization of academics claims that the mean score of its members is 112 with a standard deviation of 16. A randomly selected group of 35 members of this organization is tested, and the results reveal that the mean IQ score in this sample is 117.2. If the organization's claim is correct, what is the probability of having a sample mean of 117.2 or less for a random sample of this size? Carry your immediate computations to at least four decimal places. Round your answer to at least four decimal places. Round your answer to at least three decimal places.

0.973

Step-by-step explanation:

This is a normal distribution problem

μ = mean of population = 112

σ = standard deviation of population = 16

But for this calculation, we require the mean and standard deviation of the sample

μₓ = μ = 112

σₓ = σ/√n = 16/√35 = 2.70 (n = sample size)

We then standardize the IQ score of 117.2

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (117.2 - 112)/2.7 = 1.926

To determine the probability of having a sample mean of 117.2 or less for a random sample of this size = P(x ≤ 117.2) = P(z ≤ 1.926)

We'll use data from the normal probability table for these probabilities

P(x ≤ 117.2) = P(z ≤ 1.926) = 0.973

correct answer:

b.(19,-2)

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).