Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet.(a) What is the probability that the area covered by a can of spray paint is more than 27 square feet?(b) Suppose you want to spray paint an area of 540 square feet using 20 cans of spray paint. On average, how many square feet must each can be able to cover to spray paint all 540 square feet?(c) What is the probability that you can cover a 540 square feet area using 20 cans of spray paint?(d) If the area covered by a can of spray paint had a slightly skewed distribution, could you still calculate the probabilities in parts (a) and (c) using the normal distribution?

Step-by-step explanation:

Hello!

The variable of interest is:

X: are that can be painted with one can of spray paint.

This variable is normally distributed with mean μ= 25 feet² and standard deviation σ= 3 feet²

a. What is the probability that the area covered by a can of spray paint is more than 27 square feet?

Symbolically:

P(X≥27)

To reach the value of probability you have to use the standard normal distribution, so first you have to standardize the value of X using: Z= (X-μ)/σ

P(X≥27)= P(Z≥(27-25)/3)= P(Z≥0.67)

Now, since the Z-table has the information of cumulative probabilities: P(Zα≤Z₀)=1-α You have to do the following conversion to calculate the asked probability:

1 - P(Z<0.67)= 1 - 0.749= 0.251

b. Suppose you want to spray paint an area of 540 square feet using 20 cans of spray paint. On average, how many square feet must each can be able to cover to spray paint all 540 square feet?

If you want to paint 540 feet² using 20 cans, then each can have to paint 540/20= 27 feet² on average to cover the area.

c. What is the probability that you can cover a 540 square feet area using 20 cans of spray paint?

If you want to paint 540 feet² using the 20 cans of spray paint is the same as saying that you'll paint at least 27 feet² on average per can, symbolically:

P(X[bar]≥27)

Now you have to work with the distribution of the sample mean instead of the distribution of the variable, so for the standardization, the formula to use is Z= (X-μ)/(σ/√n).

P(X[bar]≥27)= P(Z≥(27-25)/(3/√20))= P(Z≥2.98)= 1 - P(Z≤2.98)= 1 - 0.999= 0.001

d. If the area covered by a can of spray paint had a slightly skewed distribution, could you still calculate the probabilities in parts (a) and (c) using the normal distribution?

No, if the distribution of the variable is not exactly normal, the calculations on a. and c. are not valid.

If the sample was large enough (n≥30) you could apply the central limit theorem and approximate the distribution of the sample mean to normal, if that were the case then the calculations in c. would be valid.

I hope it helps!

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

c

step-by-step explanation: