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. A. What is the probability that the diameter of a dot exceeds 0.0026 inch? B. What is the probability that a diameter is between 0.0014 and 0.0026? C. What standard deviation of diameters is needed so that the probability in part (b) is 0.995?

(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 .

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Step-by-step explanation: