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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 preferring 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) (The exact numerical value would be calculated using the formula from step 3)
Key Concept
Binomial Distribution and Cumulative Distribution Function
Explanation
The probability of at least 3 guests preferring a continental breakfast is found by subtracting the cumulative probability of 2 or fewer guests from
Question 2 (15 marks) The distance, in kilometres (km), travelled to work by the employees is under normal distribution with mean 7.5 and standard deviation 2.5. (a) What is the probability of the distance travelled to work by a randomly selected employee that: (i) Less than 11.0 km (ii) Between 5.5 km and 10.5 km (b) Find d such that 10% of the employees travel less than d kilometres to work. The entries in Table I are the probabilities that a random variable having the standard normal distribution will take on a value between 0 and z. They are given by the area of the gray region under the curve in the figure. TABLE I NORMAL-CURVE AREAS z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4648 0.4656 0.4664 0.4671 0.4678 0.4685 0.4692 0.4699 0.4706 1.9 0.4713 0.4719 0.4725 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 Also, for z = 4.0, 5.0 and 6.0, the areas are 0.49997, 0.4999997, and 0.499999999.
Solution by Steps
step 1
To calculate the probability of a randomly selected employee traveling less than 11.0 km, we first standardize the distance using the formula Z=X−μσ Z = \frac{X - \mu}{\sigma} , where X X is the distance, μ \mu is the mean, and σ \sigma is the standard deviation
step 2
For X=11.0 X = 11.0 km, μ=7.5 \mu = 7.5 km, and σ=2.5 \sigma = 2.5 km, the standardized value is Z=11.0−7.52.5 Z = \frac{11.0 - 7.5}{2.5}
step 3
Calculating the standardized value: Z=11.0−7.52.5=1.4 Z = \frac{11.0 - 7.5}{2.5} = 1.4
step 4
Using the standard normal distribution table, we find the probability for Z≤1.4 Z \leq 1.4
step 5
The table value for Z=1.4 Z = 1.4 is approximately 0.4192. This is the probability that Z Z is less than 1.4
step 6
The probability that a randomly selected employee travels less than 11.0 km is the same as the probability that Z Z is less than 1.4
Answer
The probability that a randomly selected employee travels less than 11.0 km is approximately 0.4192.
Key Concept
Standardizing a normal distribution
Explanation
By standardizing the distance to a Z-score, we can use the standard normal distribution table to find the probability of an employee traveling less than a certain distance.
step 1
To calculate the probability of the distance being between 5.5 km and 10.5 km, we standardize both distances using the formula Z=X−μσ Z = \frac{X - \mu}{\sigma}
step 2
For X1=5.5 X_1 = 5.5 km and X2=10.5 X_2 = 10.5 km, μ=7.5 \mu = 7.5 km, and σ=2.5 \sigma = 2.5 km, we calculate Z1=5.5−7.52.5 Z_1 = \frac{5.5 - 7.5}{2.5} and Z2=10.5−7.52.5 Z_2 = \frac{10.5 - 7.5}{2.5}
step 3
Calculating the standardized values: Z1=−0.8 Z_1 = -0.8 and Z2=1.2 Z_2 = 1.2
step 4
Using the standard normal distribution table, we find the probabilities for Z1 Z_1 and Z2 Z_2
step 5
The table value for Z1=−0.8 Z_1 = -0.8 is approximately 0.2119 (since the table gives the area from 0 to Z Z , we take 0.5 - 0.2881 for the negative Z-score). The table value for Z2=1.2 Z_2 = 1.2 is approximately 0.3849
step 6
The probability of the distance being between 5.5 km and 10.5 km is P(Z2)−P(Z1) P(Z_2) - P(Z_1)
step 7
Calculating the difference: 0.3849−(0.5−0.2881) 0.3849 - (0.5 - 0.2881)
Answer
The probability that a randomly selected employee travels between 5.5 km and 10.5 km is approximately 0.1730.
Key Concept
Calculating the probability between two values in a normal distribution
Explanation
We find the probabilities of two standardized values and subtract the smaller from the larger to find the probability of the variable falling between two values.
step 1
To find d d such that 10% of the employees travel less than d d kilometers, we look for the Z-score that corresponds to the bottom 10% of the standard normal distribution
step 2
Using the standard normal distribution table, we find the Z-score that has approximately 0.1000 to its left. This corresponds to a Z-score of approximately -1.28
step 3
We use the Z-score formula in reverse to find d d : d=μ+Z⋅σ d = \mu + Z \cdot \sigma
step 4
For μ=7.5 \mu = 7.5 km, σ=2.5 \sigma = 2.5 km, and Z=−1.28 Z = -1.28 , we calculate d d
step 5
Calculating d d : d=7.5+(−1.28)⋅2.5 d = 7.5 + (-1.28) \cdot 2.5
Answer
The value of d d such that 10% of the employees travel less than d d kilometers is approximately 4.8 km.
Key Concept
Finding a percentile in a normal distribution
Explanation
We use the Z-score associated with the desired percentile and the properties of the normal distribution to find the corresponding value of the random variable.
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