Question

The diameter of a dot produced by a printer can be associated with a v.a.c. normally distributed. It is known that 97, 72% of the time its diameter is greater than 0.0012 inches and 0, 62% of the times its diameter is less than 0.001 inches.

a. Find the mean and variance of the diameter of the point.

b. Find the IP that the diameter of a point is greater than 0,0026 inch.

c. Find the IP that a diameter is between 0,014 and 0, 0026 inch.

d. What is the maximum variability that will be achieved so that the IP of a diameter is between 0, 0014 and 0,0026 inches is 99,5%.?

Answer 1

Solution :

Let X be the diameter of a dot produced by a printer can be associated with a v.a.c. which is normally distributed. It is known that 97.72% of the time its diameter is greater than 0.0012 inches and 0.62% of the times its diameter is less than 0.001 inches. That is we have 2.28% of the times its diameter is less than 0.001 inches. Hence we have P(X < 0.0012 ) = 0.0228 and P(X < 0.0010 ) = 0.0062.

A. But the standard normal distribution Z we get P(Z < -2) = 0.0228 and P(Z < -2.5) = 0.0062. This implies "-2" and "-2.5" are the Z scores corresponding to "0.0012" and "0.0010" respectively.

This gives, -2 = ( 0.0012-µ ) / σ and  -2.5 = ( 0.0010-µ ) / σ where µ and σ are the mean and standard deviations of X. These two equations gives us, µ - 2*σ = 0.0012 and µ - 2.5*σ = 0.0010.

Solving these two equations we get, µ = 0.002 and σ = 0.0004.

And hence mean of X ( µ ) = 0.002 and variance ( σ² ) = 0.00000016.

B. The IP that the diameter of a point is greater than 0,0026 inch is given by; P(X > 0.0026).

The Z score for 0.0026 is, Z1= (0.0026-0.002) / 0.0004 = 1.5 and hence from the Z table we get, P( Z ≤ 1.5 ) = 0.9332. So we get P( Z > 1.5) = 1 - P(Z ≤ 1.5) = 1 - 0.9332 = 0.0668.

Hence we get, P( X > 0.0026) =0.0668.

C. The IP that a diameter is between 0,014 and 0, 0026 inch is given by; P( 0.0014 ≤ X ≤ 0.0026).

The Z score for 0.0014 and 0.0026 are, Z1= (0.0026-0.002) / 0.0004 = 1.5; Z2= (0.0014-0.002) / 0.0004 = -1.5 and hence from the Z table we get, P( -1.5 ≤ Z ≤ 1.5 ) = P( Z ≤ 1.5 ) - P( Z ≤ -1.5 ) = 0.9332 - 0.0668 = 0.8664.

Hence we get, P( 0.0014 ≤ X ≤ 0.0026)=0.8664.

D. Suppose we have P( 0.0014 ≤ X ≤ 0.0026)=0.9950. To get the maximum variability we should have, P( X > 0.0026)= (1-0.9950) / 2 = 0.0025 and P( X < 0.0014 ) = 0.0025.

Let us assume µ =0.002 and we need to find maximum value of σ that satisfies the above condition.

But the standard normal distribution Z we get P(Z < -2.807) = 0.0025 and P(Z > 2.807) = 0.0025. This implies "-2.807" and "2.807" are the Z scores corresponding to "0.0014" and "0.0026" respectively.

This gives, -2.807 = ( 0.0014 - 0.0020 ) / σ and 2.807 = ( 0.0026 - 0.0020) / σ where µ and σ are the mean and standard deviations of X. These two equations gives us, 0.0020 - 2.807*σ = 0.0014 and 0.0020 + 2.807*σ = 0.0026.

Solving these two equations we get, σ = 0.00021375.

The maximum variability that will be achieved so that the IP of a diameter is between 0, 0014 and 0,0026 inches is 99,5% is, σ = 0.00021375.