Probabilities are used to determine the chances of events
The given parameters are:
- Sample size: n = 20
- Proportion: p = 85%
<h3>(a) What is the probability that 11 out of the 20 would graduate? </h3>
Using the binomial probability formula, we have:
![P(X = x) = ^nC_x p^x(1 - p)^{n -x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5EnC_x%20p%5Ex%281%20-%20p%29%5E%7Bn%20-x%7D)
So, the equation becomes
This gives
![P(x = 11) = 167960 \times (0.85)^{11} \times 0.15^{9}](https://tex.z-dn.net/?f=P%28x%20%3D%2011%29%20%3D%20167960%20%5Ctimes%20%280.85%29%5E%7B11%7D%20%5Ctimes%200.15%5E%7B9%7D)
![P(x = 11) = 0.0011](https://tex.z-dn.net/?f=P%28x%20%3D%2011%29%20%3D%200.0011)
Express as percentage
![P(x = 11) = 0.11\%](https://tex.z-dn.net/?f=P%28x%20%3D%2011%29%20%3D%200.11%5C%25)
Hence, the probability that 11 out of the 20 would graduate is 0.11%
<h3>(b) To what extent do you think the university’s claim is true?</h3>
The probability 0.11% is less than 50%.
Hence, the extent that the university’s claim is true is very low
<h3>(c) What is the probability that all 20 would graduate? </h3>
Using the binomial probability formula, we have:
![P(X = x) = ^nC_x p^x(1 - p)^{n -x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5EnC_x%20p%5Ex%281%20-%20p%29%5E%7Bn%20-x%7D)
So, the equation becomes
This gives
![P(x = 20) = 1 \times (0.85)^{20} \times (0.15\%)^0](https://tex.z-dn.net/?f=P%28x%20%3D%2020%29%20%3D%201%20%5Ctimes%20%280.85%29%5E%7B20%7D%20%5Ctimes%20%280.15%5C%25%29%5E0)
![P(x = 20) = 0.0388](https://tex.z-dn.net/?f=P%28x%20%3D%2020%29%20%3D%200.0388)
Express as percentage
![P(x = 20) = 3.88\%](https://tex.z-dn.net/?f=P%28x%20%3D%2020%29%20%3D%203.88%5C%25)
Hence, the probability that all 20 would graduate is 3.88%
<h3>(d) The mean and the standard deviation</h3>
The mean is calculated as:
![\mu = np](https://tex.z-dn.net/?f=%5Cmu%20%3D%20np)
So, we have:
![\mu = 20 \times 85\%](https://tex.z-dn.net/?f=%5Cmu%20%3D%2020%20%5Ctimes%2085%5C%25)
![\mu = 17](https://tex.z-dn.net/?f=%5Cmu%20%3D%2017)
The standard deviation is calculated as:
![\sigma = np(1 - p)](https://tex.z-dn.net/?f=%5Csigma%20%3D%20np%281%20-%20p%29)
So, we have:
![\sigma = 20 \times 85\% \times (1 - 85\%)](https://tex.z-dn.net/?f=%5Csigma%20%3D%2020%20%5Ctimes%2085%5C%25%20%5Ctimes%20%281%20-%2085%5C%25%29)
![\sigma = 20 \times 0.85 \times 0.15](https://tex.z-dn.net/?f=%5Csigma%20%3D%2020%20%5Ctimes%200.85%20%5Ctimes%200.15)
![\sigma = 2.55](https://tex.z-dn.net/?f=%5Csigma%20%3D%202.55)
Hence, the mean and the standard deviation are 17 and 2.55, respectively.
Read more about probabilities at:
brainly.com/question/15246027