Question

The amount of paint required to paint a surface with an area of 50 m^2 is normally distributed with mean 6 L and standard deviation 0.3 L.
a. If 6.2 L of paint are available, what is the probability that the entire surface can be painted?
b. How much paint is needed so that the probability is 0.9 that the entire surface can be painted.
c. What must the standard deviation be so that the probability is 0.9 that 6.2 L of paint will be sufficient to paint the entire surface?

Answer 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: the amount of paint required to paint a surface with an area of 50m².

This variable has a normal distribution with μ= 6lts and σ= 0.3lts.

a. You have 6.2 lots of paint available, with a certain probability it is possible to use all 6.2 or less to paint the surface, symbolically:

P(X≤6.2)

Next step is to standardize the value of X so that you can use the Z-table to reach the value of probability using Z= (X-μ)/σ

P(X≤6.2)= P(Z≤(6.2-6)/0.3)= P(Z≤0.67)= 0.74857

The probability that at most 6.2lts of paint will suffice to paint the entire surface is 0.74857.

b. In this case, you want to know the amount of paint, let's call it "b" that has below it a probability of 0.9, symbolically:

P(X≤b)= 0.9

Using the standard normal distribution you have to look in the table for that value that accumulates 0.9 of probability:

Z=1.283

Now you have to use this value to reverse the standardization and reach the corresponding value of X

Z= (b-μ)/σ

1.283=(b-6)/0.3

b-6= 1.283*0.3

b= 0.3849+6

b= 6.3849 ≅6.38lts

The amount of paint that has a 0.90 probability of painting the entire surface is 6.38lts.

c. Now for the same variable with the same mean, you need to find a new value for the standard deviation.

For P(X≤6.2)=0.9

Z=1.283

Z= (X-μ)/σ

Z*σ=X-μ

σ=(X-μ)/Z

σ=(6.2-6)/1.283

σ= 0.1558≅ 0.16

The standard deviation needed for an amount of paint of 6.2lts to suffice for painting the entire surface with a probability of 0.9 is σ= 0.16lts.

I hope it helps!