Question

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15, 30, and 20 min, respectively, and the standard deviations are 1, 2, and 1.5 min, respectively. What is the probability that it takes at most 70 min of machining time to produce a randomly selected component

Answer 1

Answer:

The probability is  $P(X \le 70) = 0.9685$

Step-by-step explanation:

From the question we are told that

The mean values is $M_1 = 15 \ minutes , \ M_2 = 30\ minutes, \ M_3 = 20 \ minutes$

The standard deviation is $\sigma_1 = 1 \ minutes , \ \sigma_2 = 2 \ minutes , \ \sigma = 1.5 \ minutes$

Generally the total mean is mathematically represented as

         $\= x = \sum M_i$

=>     $\= x = 15 + 30 + 20$

=>    $\= x = 65$

Generally the total variance is  mathematically represented as

     $V(x) = \sum \sigma^2_i$

=>  $V(x) = 2^2 + 1^2 + 1.5^2$

=>  $V(x) = 7.25$

Generally the total standard deviation is mathematically represented as

        $\sigma = \sqrt{V(x)}$

=>     $\sigma = \sqrt{7.25 }$

=>     $\sigma = 2.69$

Generally the probability that it takes at most 70 min of machining time to produce a randomly selected component is mathematically represented as

        $P(X \le 70) = 1 - P( X > 70 )$

Here  $P(X > 70) = P(\frac{X - \mu }{\sigma} > \frac{70 - 65}{ 2.69} )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

        $P(X > 70) = P(Z > 1.8587 )$

From the z table  

  The area under the normal curve to the right corresponding to  1.8587 is

        $P(Z > 1.8587 ) = 0.031535$

So

      $P(X \le 70) = 1 - 0.031535$

=>  $P(X \le 70) = 0.9685$