The image is kinda blurry, but from what I can gather, the notation in the attachment means
![p_X[x_k]\equiv\mathbb P(X=k)=\begin{cases}(1-p)^{k-1}p&\text{for }k=1,2,\ldots\\0&\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=p_X%5Bx_k%5D%5Cequiv%5Cmathbb%20P%28X%3Dk%29%3D%5Cbegin%7Bcases%7D%281-p%29%5E%7Bk-1%7Dp%26%5Ctext%7Bfor%20%7Dk%3D1%2C2%2C%5Cldots%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
Then
a)![p_X[x_k\le4]\equiv\mathbb P(X\le4)=\mathbb P(X=1)+\mathbb P(X=2)+\mathbb P(X=3)+\mathbb P(X=4)](https://tex.z-dn.net/?f=p_X%5Bx_k%5Cle4%5D%5Cequiv%5Cmathbb%20P%28X%5Cle4%29%3D%5Cmathbb%20P%28X%3D1%29%2B%5Cmathbb%20P%28X%3D2%29%2B%5Cmathbb%20P%28X%3D3%29%2B%5Cmathbb%20P%28X%3D4%29)
b)
c)![p_X[1\le x_k\le2]\equiv\mathbb P(1\le X\le2)=\mathbb P(X=1)+\mathbb P(X=2)](https://tex.z-dn.net/?f=p_X%5B1%5Cle%20x_k%5Cle2%5D%5Cequiv%5Cmathbb%20P%281%5Cle%20X%5Cle2%29%3D%5Cmathbb%20P%28X%3D1%29%2B%5Cmathbb%20P%28X%3D2%29)
Then
a)
b)
c)
4/10 times (7/8+9/4) 5/7
1 answer:
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Answer:
The mean is 11.5 minutes and the standard deviation is of 6.64 minutes
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The mean is:
The standard deviation is:
Arrival time of 9:18 am and a late arrival time of 9:41 am.
9:41 is 23 minutes from 9:18. So the time is uniformily distributed between 0 and 23 minutes, so a = 0, b = 23.
Mean:
Standard deviation:
The mean is 11.5 minutes and the standard deviation is of 6.64 minutes