Answer to Question #159981 in Statistics and Probability for Qianzi

Question

A company manufactures large LCD screens knows that there is no all pixels on

their screen light, even if they spend great care when making them. In a sheet of 6 ft by

10 ft that will be cut into smaller screens, they find an average of 4.8 blank pixels. They

believe that the occurrences of blank pixels are independent. Their warranty policy

states that they will replace any screen sold that showed more than 3 blank pixels.

i. What is the mean number of blank pixels per square foot?

ii. What is the standard deviation of blank pixels per square foot?

iii. What is the probability that a 2 ft by 3 ft screen will have at least two

defects?

iv. What is the probability that a 2 ft by 3 ft screen will be replaced

because of blank pixels?

Answer 1

Manufacturing company that manufactures LCD screens knows that not all the pixels on their screen light even if they spend great care while making them. In sheet of 6 ft by 10ft that will be cut in to smaller screens, they find an average of 4.8 blank pixels. The occurrences’ of the pixels are independent and their warranty policy is that they guarantee replacement if the screen sold shows more than 3 blank pixels.

i. The mean of the blank pixel per sq. ft is given the average number of the blank pixels divided by the total area of the sheet which is cut in to a smaller screen.

Since it follows’ Poisson distribution the mean is represented by λ.

$λ = \frac{4.8}{6 \times 10} \\ = \frac{4.8}{60} \\ = 0.08$

Hence the mean number of blank pixel per sq ft is 0.08 pixels.

ii. The standard deviation of the blank pixel per sq ft is \sqrt{λ}.

$sd = \sqrt{λ} \\ = \sqrt{0.08} \\ = 0.2828$

Hence the standard deviation of the blank pixel per sq ft is 0.2828 pixels.

iii. We need to find the probability that a 2 ft by 3 ft screen will have at least two defects.

We know that mean number of pixel per sq ft is 0.08.

So the 2ft by 3 ft screen will have area of $2 \times 3 = 6 \;sq.ft$ and hence for the 2ft by 3ft screen the mean number of pixel is

$λ = 6 \times 0.08 \\ = 0.48 \;pixels$

Now the probability that a 2 ft by 3 ft screen will have at least two defects is

$P(X≥2) = 1 -( P(X=0) + P(X=1) ) \\ = 1 -( e^{-0.48} + \frac{e^{-0.48} \times 0.48^1}{1!}) \\ = 1 -1.48e^{-0.48} \\ = 1 -0.9158 \\ = 0.0842$

The probability that a 2 ft by 3 ft screen will have at least two defects is 0.0842.

iv. We need to find the probability that a 2 ft by 3 ft screen will be replaced because it has too many defects. The company guarantees replacement when it has more than 3 defects.

That is

$P(X≥4) = 1 -( P(X=0) + P(X=1) + P(X=2) + P(X=3) ) \\ = 1 -( \frac{e^{-0.48} \times 0.48}{0!} + \frac{e^{-0.48} \times 0.48^1}{1!} + \frac{e^{-0.48} \times 0.48^2}{2!} + \frac{e^{-0.48} \times 0.48^3}{3!}) \\ = 1 -( e^{-0.48} + e^{-0.48} \times 0.48 + \frac{e^{-0.48}\times 0.2304}{2} + \frac{e^{-0.48} \times 0.1106}{6}) \\ = 1 -(0.6187 + 0.2970 + 0.0712 + 0.0114) \\ = 1 -0.9983 \\ = 0.0017$

The probability that a 2 ft by 3 ft screen will because it has too many defects is 0.0017.