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A hostel in Country X can have 20 guests in capacity and offer guests a choice o...
Mar 26, 2024
A hostel in Country X can have 20 guests in capacity and offer guests a choice of a full English breakfast, a continental breakfast, or no breakfast. The probabilities of the choices being made are 0.45, 0.25 and 0.3 respectively. It is assumed that the breakfast choice is independent among guests. (a) On a particular morning, there are full houses in the hostel. Calculate the probability that (i) at least 3 guests prefer a continental breakfast, (ii) more than 4 and at most 6 guests prefer a full English breakfast. (b) Calculate the mean and the standard deviation for the number of guests requiring breakfast when there are 18 guests in the hostel.
Solution by Steps
step 1
Define the random variable X X as the number of guests who prefer a continental breakfast. Since the breakfast choice is independent among guests, X X follows a binomial distribution with parameters n=20 n = 20 and p=0.25 p = 0.25
step 2
Calculate the probability that at least 3 guests prefer a continental breakfast using the cumulative distribution function (CDF) of the binomial distribution. This is P(X≥3)=1−P(X≤2) P(X \geq 3) = 1 - P(X \leq 2)
step 3
Use the binomial probability formula to find P(X≤2) P(X \leq 2) : P(X≤2)=∑k=02(20k)(0.25)k(0.75)20−k P(X \leq 2) = \sum_{k=0}^{2} \binom{20}{k} (0.25)^k (0.75)^{20-k}
step 4
Calculate P(X≥3) P(X \geq 3) by subtracting P(X≤2) P(X \leq 2) from 1
Answer
P(X≥3)=1−P(X≤2) P(X \geq 3) = 1 - P(X \leq 2)
Key Concept
Binomial Distribution and Cumulative Probability
Explanation
The probability of at least 3 guests choosing a continental breakfast is found by subtracting the cumulative probability of 2 or fewer guests choosing it from 1.
step 1
Define the random variable Y Y as the number of guests who prefer a full English breakfast. Y Y follows a binomial distribution with parameters n=20 n = 20 and p=0.45 p = 0.45
step 2
Calculate the probability that more than 4 and at most 6 guests prefer a full English breakfast. This is P(5≤Y≤6)=P(Y=5)+P(Y=6) P(5 \leq Y \leq 6) = P(Y = 5) + P(Y = 6)
step 3
Use the binomial probability formula to find P(Y=5) P(Y = 5) and P(Y=6) P(Y = 6) : P(Y=5)=(205)(0.45)5(0.55)15 P(Y = 5) = \binom{20}{5} (0.45)^5 (0.55)^{15} P(Y=6)=(206)(0.45)6(0.55)14 P(Y = 6) = \binom{20}{6} (0.45)^6 (0.55)^{14}
step 4
Add P(Y=5) P(Y = 5) and P(Y=6) P(Y = 6) to get P(5≤Y≤6) P(5 \leq Y \leq 6)
Answer
P(5≤Y≤6)=P(Y=5)+P(Y=6) P(5 \leq Y \leq 6) = P(Y = 5) + P(Y = 6)
Key Concept
Binomial Probability for a Range of Values
Explanation
The probability of having more than 4 and at most 6 guests choosing a full English breakfast is the sum of the probabilities of exactly 5 and exactly 6 guests choosing it.
step 1
Define the random variable Z Z as the number of guests requiring breakfast (either full English or continental). Since the choices are independent, Z Z follows a binomial distribution with parameters n=18 n = 18 and p=0.45+0.25=0.70 p = 0.45 + 0.25 = 0.70 (the sum of the probabilities of choosing either breakfast)
step 2
Calculate the mean of Z Z using the formula for the mean of a binomial distribution: μ=n⋅p \mu = n \cdot p
step 3
Calculate the standard deviation of Z Z using the formula for the standard deviation of a binomial distribution: σ=n⋅p⋅(1−p) \sigma = \sqrt{n \cdot p \cdot (1 - p)}
Answer
Mean μ=18⋅0.70 \mu = 18 \cdot 0.70 , Standard Deviation σ=18⋅0.70⋅0.30 \sigma = \sqrt{18 \cdot 0.70 \cdot 0.30}
Key Concept
Mean and Standard Deviation of Binomial Distribution
Explanation
The mean and standard deviation for the number of guests requiring breakfast are calculated using the formulas for a binomial distribution with the sum of the probabilities of choosing either breakfast option.
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