Question

The diameter of a dot produced by a printer can be associated with a continuous random variable 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.
to. 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 inches.
c. Find the IP that a diameter is between 0.0014 and 0.0026 inches.
d. What is the maximum variability that will be achieved so that the IP of which a diameter is between
0.0014 and 0.0026 inches is 99.5%.?

Answer 1

a) P(X > 0.0012) = 0.9772

Or, P((X - )/ > (0.0012 - )/) = 0.9772

Or, P(Z > (0.0012 - )/) = 0.9772

Or, P(Z < (0.0012 - )/) = 0.0228

Or, (0.0012 - )/ = -2

Or, = 0.0012 + 2

P(X < 0.001) = 0.0062

Or, P((X - )/ < (0.001 - )/) = 0.0062

Or, P(Z < (0.001 - )/) = 0.0062

Or, (0.001 - )/ = -2.50

Or, = 0.001 + 2.50

Now we will get,

0.0012 + 2 = 0.001 + 2.50

Or, 0.5 = 0.0002

Or, = 0.0004

Or, = 0.00000016

= 0.001 + 2.5 * 0.0004 = 0.002

b) P(X > 0.0026)

= P((X - )/ > (0.0026 - )/)

= P(Z > (0.0026 - 0.002)/0.0004)

= P(Z > 1.5)

= 1 - P(Z < 1.5)

= 1 - 0.9332

= 0.0668

c) P(0.0014 < X < 0.0026)

= P((0.0014 - )/ < (X - )/ < (0.0026 - )/)

= P((0.0014 - 0.002)/0.0004 < Z < (0.0026 - 0.002)/0.0004)

= P(-1.5 < Z < 1.5)

= P(Z < 1.5) - P(Z < -1.5)

= 0.9332 - 0.0668

= 0.8664

d) P(0.0014 < X < 0.0026) = 0.995

Or, P((0.0014 - )/ < (X - )/ < (0.0026 - )/) = 0.995

Or, P((0.0014 - 0.002)/ < Z < (0.0026 - 0.002)/) = 0.995

Or, P(-0.0006/ < Z < 0.0006/) = 0.995

Or, P(Z < 0.0006/) - P(Z < -0.0006/) = 0.995

Or, P(Z < 0.0006/) - (1 - P(Z < 0.0006/)) = 0.995

Or, 2P(Z < 0.0006/) = 1.995

Or, P(Z < 0.0006/) = 0.9975

Or, 0.0006/ = 2.81

Or, = 0.0002

Or, = 0.00000004