Question

Suppose that the demand for a company’s product in weeks 1, 2, and 3 are each normally distributed and the mean demand during each of these three weeks is 50, 45, and 65, respectively. Suppose the standard deviation of the demand during each of these three weeks is known to be 10, 5, and 15, respectively. It turns out that if we can assume that these three demands are probabilistically independent then the total demand for the three week period is also normally distributed. And, the mean demand for the entire three week period is the sum of the individual means. Likewise, the variance of the demand for the entire three week period is the sum of the individual weekly variances. A. Suppose that the company currently has 180 units in stock, and it will not be receiving any further shipments from its supplier for at least 3 weeks. What is the probability that the companywill run out of units?B. How many units should the company currently have in stock so that it can be 98% certain of not running out during this three-week period? Again, assume that it won't receive any more shipments during this period.

Answer 1

Answer:

A

  $P(X > 180 ) =0.14253$

B

  $k = 204$

Step-by-step explanation:

From the question we are told that

  The  first mean is  $\mu_1 = 50$

  The second mean is $\mu_2 = 45$

  The third mean is $\mu_3= 65$

 The first standard deviation is $\sigma_1 = 10$

 The  second  standard deviation is  $\sigma_2 = 5$

  The  third standard deviation is $\sigma _3 = 15$

    The  number of unites in the stock is  x =  180 unites

Generally the first variance is mathematically represented as

       $\sigma^2_1 = 10^2$

=>   $\sigma^2_1 = 100$

The  second  variance is  

      $\sigma^2_2 = 5^2$

=>   $\sigma^2_2 = 25$

The  third variance is

    $\sigma^2_3 = 15^2$

=>$\sigma^2_3 = 225$

Generally the total mean is evaluated as

     $\mu =\mu_1 + \mu_2 + \mu_3$

    $\mu =50 +45 + 65$

     $\mu =160$

The  total variance is evaluated as

   $\sigma^2 = \sigma^2 _1 + \sigma^2 _2 + \sigma^2 _3$

   $\sigma^2 = 100 + 25 + 225$

   $\sigma^2 = 350$

The  standard deviation is  mathematically represented as

   $\sigma = \sqrt{350}$

    $\sigma = 18.708$

The probability that the company will run out of units is mathematically represented as

           $P(X > 180 ) = P(\frac{X - \mu }{\sigma } > \frac{ 180 - 160}{18.708} )$

            $P(X > 180 ) = P(Z > 1.069 )$

From the z-table  

           P(Z >  1.069  )  =  0.14253

So  

    $P(X > 180 ) =0.14253$

The number of unites the company needs to have in stock in order to 98% certain of not running out during this three-week period is mathematically represented as

           $P(X < k ) = P( \frac{X - \mu }{\sigma} > \frac{k - \mu }{ \sigma } ) = 0.98$

           $P(X < k ) = P( \frac{X - \mu }{\sigma} > \frac{k - 160 }{ 18.708} ) = 0.98$

From the z -table  table  the value corresponding to 98% of the area under the normal distribution curve is 2.33

 So  

        $\frac{k - 160 }{ 18.708} = 2.33$

=>   $k = 204$