Respuesta :
Answer:
[tex]a.\ \ \ \mu_T=5345, \ \ \sigma_T=600[/tex]
[tex]b.\ \ K=$5,995[/tex]
[tex]C.\ \ \mu_B=6360, \ median(B)=6360 ,\ \ \sigma_B=800[/tex]
Step-by-step explanation:
The expense function is expressed as [tex]T=15X+2300[/tex] where X is the weight of collected iron.
-This function is normal distributed as [tex]X \sim \mathcal{N}(\mu,\,\sigma^{2})[/tex], [tex]\mu_x=203, \sigma_x=40[/tex]
-We use the formula for normal distribution to calculate the mean as:
[tex]=\mu_T=15\mu_x+2300,\ \mu_x=203\\\\=15\times 203+2300\\\\=5345[/tex]
#We then calculate the standard deviation as:
[tex]Var(T)=\sigma_T^2\\\\=15^2\times \sigma_x^2\\\\=225\times 40^2\\\\=360000\\\sigma_T=\sqrt{\sigma_T^2}\\\\\therefore \sigma_T=\sqrt{360000}=600[/tex]
b. From a above, we know that the cost, T is normally distributed with mean=5345 and standard deviation=600, [tex]T\sim \mathcal{N}(5345,\,\ 600^2)[/tex]
#Given a 13.8% chance that the total daily expense of the business is more than $K, we calculate K as:
[tex]T \sim \mathcal{N}(5345,\,360000)\\\\\\\\Z=\frac{T-5345}{600} \sim \mathcal{N}(0,\,1)\\=P(T>K)=0.138\\\\\\P(T\leq K)=1-0.138=0.862\\\\P(Z\leq Z^*)=0.8620\\=>Z^*\approx 1.08935\\\\\frac{T^*-5345}{600}=1.08395\\\\T^*=1.08395\times 600+5345\\\\=5995.37\approx5995[/tex]
Hence, the value of K is approximately $5995
c. Now given that each kilo or iron collected can be sold for $35, the mean, sd and median of the daily profit is calculated as:
Let B be the profit made:
[tex]B=35X-T\\\\=20X-2300[/tex]
#Mean profit is calculated as:
[tex]\mu_B=20\mu_x-2300, \ \mu_x=203\\\\=20\times 203-2300\\\\=6360[/tex]
#In a normal distribution:
[tex]mean=mode=median\\\\\\\therefore median(B)=6360[/tex]
#The standard deviation is calculated as:
[tex]\sigma_B=\sqrt{\sigma_B^2}\\\\=\sqrt{20^2\sigma_x^2}\\\\=\sqrt{20^2\times 40^2}\\\\=800\\\\\therefore \sigma_B=800[/tex]
d. A government allowance is applicable if theres 10+% of a negative profit;
[tex]B \sim \mathcal{N}(6360,\,\ 800^{2})\\\\\therefore Z=\frac{B\leq -6360}{800} \sim \mathcal{N}(0,1)\\\\P(Z\leq Z^*)=0.1, Z^*\approx-1.2816\\\\\frac{B^*-6360}{800}=-1.2816\\\\B^*=-1.2816\times 800+6360\\\\=5334\\\\B=1-\frac{5334.72}{6360}=0.1612 \ or \ 16.12\%[/tex]
16.2%>10%.
Hence, he can apply for government allowance.