Question

Estimate the indicated probability by using the normal approximation as an approximation to the binomial distribution. (PROBLEMS 4 & 5) 4. Two percent of hair dryers produced in a certain plant are defective. Estimate the probability that of 10,000 randomly selected hair dryers, at least 219 are defective.
5. In one county, the conviction rate for speeding is 85%. Estimate the probability that of the next 100 speeding summonses issued, there will be at least 90 convictions.

Answer 1

Answer:

4

  $P(X \ge 219 ) = 0.093$

5

 $P(X \ge 219 ) = 0.1038$

Step-by-step explanation:

From the question we are told that  

    The percentage of hair dryers that are defective is p=2%  $= 0.02$

      The  sample size is  $n =10000$

       The  random number is  x = 219

The mean of this data set is evaluated as

     $\mu = n* p$

substituting values

     $\mu = 10000 * 0.02$

    $\mu =200$

The standard deviation is evaluated as

      $\sigma = \sqrt{[\mu (1 -p)]}$

substituting values  

      $\sigma = \sqrt{[200 (1 -0.02)]}$

     $\sigma = 14$

Since it is  a one tail test the degree of freedom is  

      df =  0.5

So  

   $x_l = x- 0.5$

   $x = 219 - 0.5$

    $x = 218.05$

Now applying normal approximation

    $P(X \ge 219 ) = P(z > \frac{x - \mu}{\sigma} )$

substituting values  

    $P(X \ge 219 ) = P(z > \frac{218.5 - 200}{14} )$

    $P(X \ge 219 ) = P(z > 1.32} )$

From the z-table  

    $P(X \ge 219 ) = 0.093$

Considering Question 5  

The  random number is  x = 90

      The mean is  $\mu = n * p$

Where  n = 100

   and  p  =  0.85

So  

     $\mu = 0.85 *100$    

     $\mu = 85$

The standard deviation is  evaluated as

     $\sigma = \sqrt{[\mu (1 -p)]}$

substituting values  

      $\sigma = \sqrt{[85 (1 -0.85)]}$        

     $\sigma = 3.5707$      

Since it is  a one tail test the degree of freedom is  

      df =  0.5

Now applying normal approximation

    $P(X \ge 90 ) = P(z > \frac{x - \mu}{\sigma} )$

   substituting values  

    $P(X \ge 219 ) = P(z > \frac{89.5 - 85}{3.5707} )$

    $P(X \ge 219 ) = P(z > 1.2603} )$

From the z-table  

   $P(X \ge 219 ) = 0.1038$