The amount of coffee that people drink per day is normally distributed with a mean of 15 ounces and a standard deviation of 4.5 ounces. 10 randomly selected people are surveyed. Round all answers to 4 decimal places where possible. What is the distribution of X ? X ~ N( 15 , 4.5 ) What is the distribution of ¯ x ? ¯ x ~ N( 15 , ) What is the probability that one randomly selected person drinks between 14.5 and 16.5 ounces of coffee per day? For the 10 people, find the probability that the average coffee consumption is between 14.5 and 16.5 ounces of coffee per day. For part d), is the assumption that the distribution is normal necessary? YesNo Find the IQR for the average of 10 coffee drinkers. Q1 = ounces Q3 = ounces IQR: ounces

Step-by-step explanation:

Hello!

The variable of interest is

X: the amount of coffee a person drinks per day. (ounces)

This variable has a normal distribution with μ= 15 ounces and σ=4.5 ounces.

10 randomly selected persons where surveyed.

a) X~N(μ;σ²) ⇒ X~N(15;20.25)

b) The sample means has the same distribution as the original variable except that the population variance is affected by the sample size:

X[bar]~N(μ;σ²/n) ⇒ X[bar]~N(15;20.25/10) ⇒ X[bar]~N(15;2.025)

c)

You need to find the probability that one random person drinks between 14.5 and 16.5 ounces of coffee per day, symbolically:

P(14.5 ≤ X ≤ 16.5)

For this part, since you need to calculate the probability that one person drinks between 14.5 and 16.5 ounces of coffee daily out of the whole population of people that drink coffee, you have to work using the distribution of the variable:

X~N(15;20.25)

To reach the correspondent probability you have to first standardize both bonds of the interval using Z=(X-μ)/δ~N(0;1):

P(14.5 ≤ X ≤ 16.5)

P(X ≤ 16.5) - P(X ≤ 14.5)

P(Z≤ (16.5-15)/4.5) - P(Z≤ (14.5-15)/4.5)

P(Z≤ 0.33) - P(Z≤ -0.11)= 0.62930 - 0.45620= 0.1731

d)

In this item, you need to calculate the probability that one out of the ten people surveyed drinks on average between 14.5 and 16.5 ounces of coffee.

P(14.5 ≤ X[bar] ≤ 16.5)

To calculate the correspondent probability you have to work using the distribution od the sample mean: X[bar]~N(15;2.025) and the standard normal: Z=(X[bar]-μ)/(δ/√n)~N(0;1)

P(X[bar] ≤ 16.5) - P(X[bar] ≤ 14.5)

P(Z≤ (16.5-15)/1.42) - P(Z≤ (14.5-15)/1.42)

P(Z≤ 1.06) - P(Z≤ -0.35)= 0.85543 - 0.36317= 0.49226

e)

Yes. Without the assumption that the variable had a normal distribution, you wouldn't be able to use the standard normal distribution to calculate the asked probabilities.

There is a possibility to apply the Central Limit Theorem to approximate the sampling distribution to normal if the sample was large enough (n ≥30) which is not the case.

f)

You need to find the IQR for the average consumption of coffee for a sample of 10 persons.

For this, you have to work using the distribution of the sample mean X[bar]~N(15;2.025)

The 1st quartile is the value that divides the bottom 25% of the distribution from the top 75%, symbolically:

P(X[bar]≤x[bar]₀)= 0.25

First, you have to determine the value of Z that accumulates 0.25 of probability:

P(Z≤z₀)= 0.25

z₀= -0.674

Now using the formula Z=(X[bar]-μ)/(δ/√n), you have to "reverse" the standardization, i.e. clear the value of X[bar]:

z₀=(x[bar]₀-μ)/(δ/√n)

-0.674=(x[bar]₀-15)/1.42

-0.674*1.42= (x[bar]₀-15)

x[bar]₀= (-0.674*1.42)+15

1st Quartile x[bar]₀= 14.0429 ounces

The third quartile is the value that divides the bottom 75% of the distribution from the top 25% of it:

P(X[bar]≤x[bar]₀)= 0.75

As before, the first step is to determine the corresponding value of Z:

P(Z≤z₀)= 0.75

z₀= 0.674

z₀=(x[bar]₀-μ)/(δ/√n)

0.674= (x[bar]₀-15)/1.42

0.674*1.42= x[bar]₀-15

x[bar]₀= (0.674*1.42)+15

3rd Quartile x[bar]₀= 15.9571

The IQR is the difference between Q3 and Q1:

IQR= 15.9571 - 14.0429= 1.9142

I hope this helps!

11.2

step-by-step explanation:

when doing tangent the tangent of the angle is equal to the opposite side from the angle over the adjacent side to the angle.

tan43 = x / 12

move 12 over by multiplying each side by 12

12tan43 = x

x = about 11.2