Answer: $53.95
Step-by-step explanation:
Given: [tex]\mu=0.75[/tex] pounds
[tex]\sigma= 0.2[/tex] pounds
n= 50
Two tailed critical z-value for 90% confidence level = 1.645
Let x be the total amount of food that a cafeteria should have on hand to be 90 percent confident that it will not run out of food .
z-score = [tex]\dfrac{x-\mu}{\sigma}[/tex] [tex]=\dfrac{x-0.75}{0.2}[/tex]
Then, [tex]\dfrac{x-0.75}{0.2}=1.645[/tex]
[tex]x-0.75=1.645\times0.2\\\\\Riightarrow\ x-0.75=0.329\\\\\Rightarrow\ x= 0.329+0.75\\\\\Rightarrow\ x= 1.079[/tex]
Foe 50 students, total amount of food = 1.079 x 50 = $53.95
hence, the cafeteria should have food of amount $53.95 so that it is be 90 percent confident that it will not run out of food.
