Answer:
A
[tex]P(X > 180 ) =0.14253[/tex]
B
[tex]k = 204[/tex]
Step-by-step explanation:
From the question we are told that
The first mean is [tex]\mu_1 = 50[/tex]
The second mean is [tex]\mu_2 = 45[/tex]
The third mean is [tex]\mu_3= 65[/tex]
The first standard deviation is [tex]\sigma_1 = 10[/tex]
The second standard deviation is [tex]\sigma_2 = 5[/tex]
The third standard deviation is [tex]\sigma _3 = 15[/tex]
The number of unites in the stock is x = 180 unites
Generally the first variance is mathematically represented as
[tex]\sigma^2_1 = 10^2[/tex]
=> [tex]\sigma^2_1 = 100[/tex]
The second variance is
[tex]\sigma^2_2 = 5^2[/tex]
=> [tex]\sigma^2_2 = 25[/tex]
The third variance is
[tex]\sigma^2_3 = 15^2[/tex]
=>[tex]\sigma^2_3 = 225[/tex]
Generally the total mean is evaluated as
[tex]\mu =\mu_1 + \mu_2 + \mu_3[/tex]
[tex]\mu =50 +45 + 65[/tex]
[tex]\mu =160[/tex]
The total variance is evaluated as
[tex]\sigma^2 = \sigma^2 _1 + \sigma^2 _2 + \sigma^2 _3[/tex]
[tex]\sigma^2 = 100 + 25 + 225[/tex]
[tex]\sigma^2 = 350[/tex]
The standard deviation is mathematically represented as
[tex]\sigma = \sqrt{350}[/tex]
[tex]\sigma = 18.708[/tex]
The probability that the company will run out of units is mathematically represented as
[tex]P(X > 180 ) = P(\frac{X - \mu }{\sigma } > \frac{ 180 - 160}{18.708} )[/tex]
[tex]P(X > 180 ) = P(Z > 1.069 )[/tex]
From the z-table
P(Z > 1.069 ) = 0.14253
So
[tex]P(X > 180 ) =0.14253[/tex]
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
[tex]P(X < k ) = P( \frac{X - \mu }{\sigma} > \frac{k - \mu }{ \sigma } ) = 0.98[/tex]
[tex]P(X < k ) = P( \frac{X - \mu }{\sigma} > \frac{k - 160 }{ 18.708} ) = 0.98[/tex]
From the z -table table the value corresponding to 98% of the area under the normal distribution curve is 2.33
So
[tex]\frac{k - 160 }{ 18.708} = 2.33[/tex]
=> [tex]k = 204[/tex]
