Answer:
Step-by-step explanation:
-5 = -5x
1 = x
Answer:
![P(x\geq 3)=0.3233](https://tex.z-dn.net/?f=P%28x%5Cgeq%203%29%3D0.3233)
Step-by-step explanation:
If the number of defects in poured metal follows a Poisson distribution, the probability that x defects occurs is:
![P(x)=\frac{e^{-m}*(m)^{x}}{x!}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7Be%5E%7B-m%7D%2A%28m%29%5E%7Bx%7D%7D%7Bx%21%7D)
Where x is bigger or equal to zero and m is the average. So replacing m by 2, we get that the probability is equal to:
![P(x)=\frac{e^{-2}*(2)^{x}}{x!}](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7Bx%7D%7D%7Bx%21%7D)
Finally, the probability that there will be at least three defects in a randomly selected cubic millimeter of this metal is equal to:
![P(x\geq 3)=1-p(x\leq 2)\\](https://tex.z-dn.net/?f=P%28x%5Cgeq%203%29%3D1-p%28x%5Cleq%202%29%5C%5C)
Where ![P(x\leq 2)=P(0)+P(1)+P(2)](https://tex.z-dn.net/?f=P%28x%5Cleq%202%29%3DP%280%29%2BP%281%29%2BP%282%29)
So, P(0), P(1) and P(2) are equal to:
![P(0)=\frac{e^{-2}*(2)^{0}}{0!}=0.1353\\P(1)=\frac{e^{-2}*(2)^{1}}{1!}=0.2707\\P(2)=\frac{e^{-2}*(2)^{2}}{2!}=0.2707](https://tex.z-dn.net/?f=P%280%29%3D%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7B0%7D%7D%7B0%21%7D%3D0.1353%5C%5CP%281%29%3D%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7B1%7D%7D%7B1%21%7D%3D0.2707%5C%5CP%282%29%3D%5Cfrac%7Be%5E%7B-2%7D%2A%282%29%5E%7B2%7D%7D%7B2%21%7D%3D0.2707)
Finally,
and
are equal to:
![P(x\leq 2)=0.1353+0.2707+0.2707=0.6767\\P(x\geq 3)=1-0.6767=0.3233](https://tex.z-dn.net/?f=P%28x%5Cleq%202%29%3D0.1353%2B0.2707%2B0.2707%3D0.6767%5C%5CP%28x%5Cgeq%203%29%3D1-0.6767%3D0.3233)
Answer:
C.I. at 95% is as C.I.[0.7504 to 0.8162]
Step-by-step explanation:
Given:
Total number n=600
and Required No of students=x=470
To Find:
Determine a 95% confidence interval .
Solution:
Using normal distribution table for Z,
when 95 % C.I. Z=1.96
Now calculate ,
Population proportion,p=x/n
p=470/600
p=0.7833
Now calculate margin of error
MOE=Z*Sqrt[p(1-p)/n]
MOE=1.96*Sqrt[(0.7833*0.21667)/600]
=1.96*Sqrt(0.00028286)
=1.96*0.01681
=0.0329
So
The Boundaries will be as follows:
1)p-MOE
=0.7833-0.0329
=0.7504
2)p+MOE
=0.7833+0.0329
=0.8162
Answer:
21.16
Step-by-step explanation:
Starting from the theory we have the following equation:
![fi*P(x](https://tex.z-dn.net/?f=fi%2AP%28x%3Cc-1%29%20%3D%200.99)
Using the data supplied in the exercise, we have subtracting the mean and dividing by the standard deviation:
![P( z \leq \frac{c-1-12}{3.5}) =0.99/fi](https://tex.z-dn.net/?f=P%28%20z%20%5Cleq%20%5Cfrac%7Bc-1-12%7D%7B3.5%7D%29%20%3D0.99%2Ffi)
solving for "c", knowing that fi is a tabulating value:
![\frac{c-13}{3.5}=0.99/fi\\\frac{c-13}{3.5}=2.33\\c-13=2.33*3.5\\c = 8.155 +13\\c = 21.155](https://tex.z-dn.net/?f=%5Cfrac%7Bc-13%7D%7B3.5%7D%3D0.99%2Ffi%5C%5C%5Cfrac%7Bc-13%7D%7B3.5%7D%3D2.33%5C%5Cc-13%3D2.33%2A3.5%5C%5Cc%20%3D%208.155%20%2B13%5C%5Cc%20%3D%2021.155)
therefore the value of c is equal to 21.16
(I) The 15 represents the money earned for each ticket sold
In other words the 15 is the cost of each ticket.
(II) The 700 represents the costs inquired when organizing the dance
In other words...how much it costs to throw the dance.
<u>Explanation</u>
Since t is the number of tickets and the function shows profit we know that every step on the left has some aspect to do with the profit....for ever ticket (t) sold there is an addition of 15 dollars...or for every 1 ticket sold $15 is earned....or each ticket costs 15 dollars
The -700 shows that 700 dollars must be subtracted from the total income from selling tickets....since this is looking for profit which is income-costs and we have already determined that 15t represents the income we know that the 700 must represent the costs of throwing the dance.