Q1 a) A company produces paper and it is known that the production follows a normal distribution with a mean of 500 and a standard deviation of 80.What is the probability that x is less than 400?

b) Let x be a random variable with a mean of 70 and standard deviation of 8. If x is normally distributed , what is the probability that x is at least 50?

c) A population has a mean of 4 and a standard deviation of 5 . What is the probability that the random variable x is greater than 4.5 ? If x is normally distributed.

Step-by-step explanation:

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