Answer:
(a) 3.55
(b) 3.45 and 1.86
(c) 0.25
(d) 0.016
Step-by-step explanation:
The random variable X denotes the number of people in a group (party) at a restaurant.
(a)
The formula to compute the mean is:
[tex]\mu=\sum {x\cdot P(X=x)}[/tex]
Consider the Excel sheet attached.
The mean is, 3.55.
(b)
The formula to compute the variance is:
[tex]\sigma^{2}=[\sum {x^{2}\cdot P(X=x)}]-(\mu)^{2}[/tex]
Consider the Excel sheet attached.
Compute the variance as follows:
[tex]\sigma^{2}=[\sum {x^{2}\cdot P(X=x)}]-(\mu)^{2}\\\\=16.05-(3.55)^{2}\\\\=3.4475\\\\\approx 3.45[/tex]
The variance is, 3.45.
Compute the standard deviation as follows:
[tex]\sigma=\sqrt{\sigma^{2}}\\\\=\sqrt{3.45}\\\\=1.85742\\\\\approx 1.86[/tex]
The standard deviation is, 1.86.
(c)
Compute the probability that the next party will be over 4 people as follows:
[tex]P(X>4)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\\\=0.10+0.05+0.05+0.05\\\\=0.25[/tex]
Thus, the probability that the next party will be over 4 people is 0.25.
(d)
Compute the probability that the next three parties will each be over 4 people as follows:
It is provided that the three parties are independent.
P (Next 3 parties will be each over 4) = [P (X > 4)]³
[tex]=(0.25)^{3}\\=0.015625\\\approx 0.016[/tex]
Thus, the probability that the next three parties will each be over 4 people is 0.016.
