Question

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?

Answer 1

Answer:

(a) 0.06681

(b) 0.86638

(c)  $\sigma$ = 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.;   $\mu$ = 0.002 inch        and        $\sigma$ = 0.0004

Also,   Z = $\frac{X -\mu}{\sigma}$ ~ N(0,1)

(a) Let X = diameter of a dot

    P(X > 0.0026 inch) = P( $\frac{X -\mu}{\sigma}$ > $\frac{0.0026 -0.002}{0.0004}$ ) = 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( $\frac{X -\mu}{\sigma}$ < $\frac{0.0026 -0.002}{0.0004}$ ) = P(Z < 1.5) = 0.93319

    P(X <= 0.0014) = P( $\frac{X -\mu}{\sigma}$ <= $\frac{0.0014 -0.002}{0.0004}$ ) = 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( $\frac{0.0014 -0.002}{\sigma}$ < $\frac{X -\mu}{\sigma}$ < $\frac{0.0026 -0.002}{\sigma}$ ) = 0.995

    P( $\frac{ -0.0006}{\sigma}$ < Z < $\frac{0.0006}{\sigma}$ ) = 0.995

    P(Z < $\frac{0.0006}{\sigma}$ ) - P(Z <= $\frac{-0.0006}{\sigma}$ ) = 0.995

    P(Z < $\frac{0.0006}{\sigma}$ ) - (1 - P(Z < $\frac{0.0006}{\sigma}$ ) ) = 0.995

     2 * P(Z < $\frac{0.0006}{\sigma}$ ) - 1 = 0.995

           P(Z < $\frac{0.0006}{\sigma}$ ) = 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.;

                  $\frac{0.0006}{\sigma}$ = 2.81      ⇒  $\sigma$ = 0.000214

Therefore, 0.000214 standard deviation of diameters is needed so that the probability in part (b) is 0.995 .

Answer 2

The probabilities of obtaining a given diameter is found from the z-table,

given that the dot produced by the printer are normally distributed.

• A. The probability that the diameter of a dot exceeds 0.0026 inch is 0.0668

• B. The probability that the diameter is between 0.0014 and 0.0026 inch is 0.8664

• C. The standard deviation needed for a probability of 0.995 is 2.135 × 10⁻⁴

Reasons:

The mean diameter, μ = 0.002

The standard deviation, σ = 0.0004

A. The probability that the diameter exceeds 0.0026 inch

Solution;

$\displaystyle z-score,\ Z= \mathbf{\dfrac{x-\mu }{\sigma }}$

At x = 0.0026 inch, we have;

$\displaystyle Z=\dfrac{0.0026-0.002 }{0.0004 } = 1.5$

P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668

The probability that the diameter of a dot exceeds 0.0026 inch = 0.0668

B. The probability that the diameter is less than 0.0026 inch = 0.9332

The z-score for a diameter of x = 0.0014 inch is given as follows;

$\displaystyle Z=\mathbf{\dfrac{0.0014-0.002 }{0.0004 }} = -1.5$

P(Z < -1.5) = 0.0668

Therefore, the probability that the diameter is between 0.0014 and 0.0026 inch is given as follows;

P(0.0014 < x < 0.0026) = 0.9332 - 0.0668 = 0.8664

The probability that the diameter is between 0.0014 and 0.0026 = 0.8664

C. For the probability in part (b) to be 0.995, we have;

For a probability of 0.995, the z-score ≈ 2.575

$\displaystyle P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right)- P\left(Z < \dfrac{0.0014-0.002 }{\sigma } \right)= 0.995$

Therefore;

$\displaystyle \mathbf{ P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right)} = 0.995 + \frac{1 - 0.995}{2} = 0.9975$

From the z-table, we get;

$\displaystyle P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right) = 0.9975$

The z-score with a probability of 0.9975 = 2.81

Which gives;

$\displaystyle \left( \dfrac{0.0026-0.002 }{\sigma } \right) = 2.81$

$\displaystyle \sigma = \left( \dfrac{0.0026-0.002 }{2.81} \right) = \mathbf{2.135 \times 10^{-4}}$

The standard deviation of the diameters needed so that the probability in part (b) is 0.995 is 2.135 × 10⁻⁴

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