Question

A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed into a can is known to have a normal distribution with a mean of 5 milliliters (ml) and a standard deviation of 0.4 ml. Answer the following questions based on this information. Find the dye amount that represents the 9th percentile of the distribution. 4.464 ml 4.836 ml 4.936 ml 4.964 ml 5.536 ml

Answer 1

Answer:

The 9th percentile is 4.464ml

Explanation:

Hello!

Given the variable X: Amount of dye dispensed into a paint can.

With a normal distribution, mean μ= 5ml and standard deviation δ= 0.4ml

You need to calculate the 9th percentile of the distribution.

The percentile is a measure of the position that indicates the number that separates the distribution or data set in a percentage of interest. In this case, the 9th percentile is the value of the distribution that separates the bottom 9% from the top 91%.

Symbolically:

P(X≤x₀)= 0.09

x₀ represents the value of the percentile.

The best way to calculate this value is by using the standard normal distribution since it is already tabulated, and then "translate" the Z-value to a value of the variable.

Under the standard normal distribution you have to look for:

P(Z≤z₀)= 0.09

The value marks the bottom 9% of the distribution, this means that you'll find it in the left tail of it. Remember the mean of the standard normal distribution is zero, so all values under the mean will be negative. Using the left entry of the Z-table you have to look for 0.09 in the body of the table and reach the margins to find the corresponding value:

z₀= -1.34

Now using the formula of the distribution you can "translate" the Z-value in terms of the variable of interest

Z=(X-μ)/ δ ~N(0;1)

z₀=(x₀-μ)/ δ

z₀*δ=x₀-μ

x₀=(z₀*δ)+μ

x₀=(-1.34*0.4)+5

x₀= 4.464

The 9th percentile is 4.464ml

I hope you have a SUPER day!