Answer:
(a) 0.06681
(b) 0.86638
(c) = 0.000214
Step-by-step explanation:
We are given that the diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch and a standard deviation of 0.0004 inch i.e.; = 0.002 inch and = 0.0004
Also, Z = ~ N(0,1)
(a) Let X = diameter of a dot
P(X > 0.0026 inch) = P( > ) = P(Z > 1.5) = 1 - P(Z <= 1.5)
= 1 - 0.93319 = 0.06681
(b) P(0.0014 < X < 0.0026) = P(X < 0.0026) - P(X <= 0.0014)
P(X < 0.0026) = P( < ) = P(Z < 1.5) = 0.93319
P(X <= 0.0014) = P( <= ) = P(Z <= -1.5) = 1 - P(Z <= 1.5)
= 1 - 0.93319 = 0.06681
Therefore, P(0.0014 < X < 0.0026) = 0.93319 - 0.06681 = 0.86638 .
(c) P(0.0014 < X < 0.0026) = 0.995
P( < < ) = 0.995
P( < Z < ) = 0.995
P(Z < ) - P(Z <= ) = 0.995
P(Z < ) - (1 - P(Z < ) ) = 0.995
2 * P(Z < ) - 1 = 0.995
P(Z < ) = 0.9975
On seeing the z table we observe that at critical value of x = 2.81 we get the probability area of 0.9975 i.e.;
= 2.81 ⇒ = 0.000214
Therefore, 0.000214 standard deviation of diameters is needed so that the probability in part (b) is 0.995 .