The answer would be 1.25.
Answer:
a. 0.7291
b. 0.9968
c. 0.7259
Step-by-step explanation:
a. np and n(1-p) can be calculated as:

#Both np and np(1-p) are greater than 5, hence, normal approximation is most appropriate:

#Define Y:
Y~(11.04,5.7408)

Hence, the probability of 12 or fewer is 0.8291
b. The probability that 5 or more fish were caught.
#Using normal approximation:

Hence, the probability of catching 5+ is 0.9968
c. The probability of between 5 and 12 is calculated as;
-From b above
and a ,
=0.7291

Hence, the probability of between 5 and 12 is 0.7259
Answer:
The probability that the restaurant can accommodate all the customers who do show up is 0.3564.
Step-by-step explanation:
The information provided are:
- At 7:00 pm the restaurant can seat 50 parties, but takes reservations for 53.
- If the probability of a party not showing up is 0.04.
- Assuming independence.
Let <em>X</em> denote the number of parties that showed up.
The random variable X follows a Binomial distribution with parameters <em>n</em> = 53 and <em>p</em> = 0.96.
As there are only 50 sets available, the restaurant can accommodate all the customers who do show up if and only if 50 or less customers showed up.
Compute the probability that the restaurant can accommodate all the customers who do show up as follows:
![P(X\leq 50)=1-P(X>50)\\=1-P(X=51)-P(X=52)-P(X=53)\\=1-[{53\choose 51}(0.96)^{51}(0.04)^{53-51}]-[{53\choose 52}(0.96)^{52}(0.04)^{53-52}]\\-[{53\choose 53}(0.96)^{53}(0.04)^{53-53}]\\=1-0.27492-0.25377-0.11491\\=0.3564](https://tex.z-dn.net/?f=P%28X%5Cleq%2050%29%3D1-P%28X%3E50%29%5C%5C%3D1-P%28X%3D51%29-P%28X%3D52%29-P%28X%3D53%29%5C%5C%3D1-%5B%7B53%5Cchoose%2051%7D%280.96%29%5E%7B51%7D%280.04%29%5E%7B53-51%7D%5D-%5B%7B53%5Cchoose%2052%7D%280.96%29%5E%7B52%7D%280.04%29%5E%7B53-52%7D%5D%5C%5C-%5B%7B53%5Cchoose%2053%7D%280.96%29%5E%7B53%7D%280.04%29%5E%7B53-53%7D%5D%5C%5C%3D1-0.27492-0.25377-0.11491%5C%5C%3D0.3564)
Thus, the probability that the restaurant can accommodate all the customers who do show up is 0.3564.