Question

If a certain machine makes electrical resistors having a mean resistance of 40 ohms and a standard deviation of 2 ohms, what is the probability that a random sample of 25 of these resistors will have a combined resistance of less than 1010 ohms?

Answer 1

Answer:

The probability is  $P(\= X < 40.4) = 0.84134$

Step-by-step explanation:

From the question we are told that

   The mean is  $\mu =40 \ \Omega$

   The standard deviation is  $\sigma = 2 \ \Omega$

   The sample size is  n = 25

    The combined resistance is   $\sum x_i = 1010 \ \Omega$

Generally the sample mean is mathematically represented as

      $\= x = \frac{\sum x_i }{n}$

=>   $\= x = \frac{1010 }{25}$

=>   $\= x = 40.4 \ \Omega$

Generally the standard error of the mean is mathematically represented as

     $\sigma_{x} = \frac{\sigma }{\sqrt{n} }$

=>  $\sigma_{x} = \frac{ 2 }{\sqrt{25} }$

=>  $\sigma_{x} = 0.4$

Generally the probability that a random sample of 25 of these resistors will have a combined resistance of less than 1010 ohms is mathematically represented as

       $P(\= X < 40.4) = P( \frac{\= X - \mu }{\sigma_{x }} < \frac{ 40.4 - 40 }{0.4} )$

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

    $P(\= X < 40.4) = P( Z < 1 )$

From the z table  the area under the normal curve to the left corresponding to  1 is

     $P(\= X < 40.4) = P( Z < 1 ) = 0.84134$

     $P(\= X < 40.4) = 0.84134$