The lengths of adult males' hands are normally distributed with mean 190 mm and standard deviation is 7.4 mm. Suppose that 45 individuals are randomly chosen. Round all answers to 4 where possible. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
For the group of 45, find the probability that the average hand length is less than 189.
Find the third quartile for the average adult male hand length for this sample size.
For part b), is the assumption that the distribution is normal necessary?

a. The distribution of the sample means is normal with mean 190 mm and standard deviation 1.1031 mm.

b. The probability that the average hand length is less than 189 is P(M<189)=0.1823.

c. The third quartile for the average adult male hand length for this sample size is M_75=190.7440.

d. The assumption of normality is not necessary as the sampling distribution will tend to have a bell shaped independently of the population distribution.

Step-by-step explanation:

We have a normal distribution, with mean 190 and standard deviation 7.4.

We take samples of size n=45 from this population.

Then, the sample means will have a distribution with the following parameters:

The probability that the sample mean is less than 189 can be calculated as:

The third quartile represents the value of the sample where 75% of the data is to the left of this value. It means that:

The third quartile corresponds to a z-value of z*=0.6745.

Then, we can calculate the sample mean for the third quartile as:

The assumption of normality is not necessary as the sampling distribution will tend to have a bell shaped independently of the population distribution.

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).