Logo

AskSia

Plus

Student Question: A hostel in Country X\mathrm{X} can have 20 guests in capaci...
Mar 27, 2024
Student Question: A hostel in Country X\mathrm{X} can have 20 guests in capacity and offer guests a choice of a full English breakfast, a continental breakfast, or no breakfast. The probabilites of the choices being made are 0.45,0.250.45,0.25 and 0.3 respectively. It is assumed that the breakfast choice is independent among guests. (a) On a particular moming, there are fill houses in the hostel. Caiculate the probability that (i) at least 3 guests prefer a continental breakfast, (i) 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. Sia Response: Solution by Steps step 1 Define the random variable XX as the number of guests who prefer a continental breakfast. Since the breakfast choice is independent among guests, XX follows a binomial distribution with parameters nn =20=20 and p=0.25p=0.25. step 2 Caiculate 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≤2)=P(X \leq 2): P(X \leq 2)= sum_(k=0) (2) binom (20y(k)(20 y(k) (0.25)∗k(0.75)×(20−k)(0.25) * k(0.75) \times(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 breaidast is found by subtracting the cumulative probability of 2 or fewer guests choosing it from 1. step 1 Define the random variable YY as the number of guests who prefer a full English breakfast YY follows a binomial distribution with parameters n=20\mathrm{n}=20 and p=0.45\mathrm{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=5)=binom⁡(20)(5)P(Y=6): P(Y=5)=\operatorname{binom}(20)(5) (0.45)∧5(0.55)∧(15)P(Y=6)=binom⁡(20)(6)(0.45)∧6(0.55)∧(14)(0.45)^{\wedge 5}(0.55)^{\wedge}(15) \mathrm{P}(\mathrm{Y}=6)=\operatorname{binom}(20)(6)(0.45)^{\wedge 6}(0.55)^{\wedge}(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 quests 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 ZZ as the number of guests requiring breakfast (eilther full English or continental). Since the choices are independent, Z\mathrm{Z} follows a binomial distribution with parameters n=\mathrm{n}= 18 and p=0.45+0.25=0.70p=0.45+0.25=0.70 (the sum of the probabilities of choosing either breakfast). step 2 Calculate the mean of Z\mathrm{Z} using the formula for the mean of a binomial distribution: mu =n⋅p=n \cdot p step 3 Calculate the standard deviation of Z\mathrm{Z} using the formula for the standard deviation of a binomial distribution: sigma =(n⋅p⋅(1−p))=\sqrt{ }(n \cdot p \cdot(1-p)) Answer: Mean mu =18⋅0.70=18 \cdot 0.70, Standard Deviation sigma =(18⋅0.70⋅0.30)=\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.
Solution by Steps
step 1
Define the random variable XX as the number of guests who prefer a continental breakfast. XX follows a binomial distribution with parameters n=20n = 20 and p=0.25p = 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: 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−kP(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.
Solution by Steps
step 1
Define the random variable YY as the number of guests who prefer a full English breakfast. YY follows a binomial distribution with parameters n=20n = 20 and p=0.45p = 0.45
step 2
Calculate the probability that more than 4 and at most 6 guests prefer a full English breakfast: 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)15P(Y = 5) = \binom{20}{5} (0.45)^5 (0.55)^{15}, P(Y=6)=(206)(0.45)6(0.55)14P(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.
Solution by Steps
step 1
Define the random variable ZZ as the number of guests requiring breakfast (either full English or continental). ZZ follows a binomial distribution with parameters n=18n = 18 and p=0.45+0.25=0.70p = 0.45 + 0.25 = 0.70
step 2
Calculate the mean of ZZ using the formula for the mean of a binomial distribution: μ=n⋅p\mu = n \cdot p
step 3
Calculate the standard deviation of ZZ 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.
*al96tal40tater8eten4=4=40tan\mathrm{tan}
θ\thetatoneninsentan⁡\tanbocoA.t.a=0a=0ins1.00testsen⁡\operatorname{sen}
oltowesoentalescos\mathrm{cos}seralos400t.alss.t.cos⁡\cos
4140040.2!tollsen\mathrm{sen}4=14=1thentotstielcine
oftinetsicetastestertinsaivet.t ine?insisen
84toresint4sec4 \mathrm{sec}sienerosa10somtoes4=4==100=100
4ans10%10 \%480a.anes4x=4 x=4.30t.3an1.10==
6.tialanteraresatareresaleinestos- 164elve
of43043ane?4.04:4:470+ros4×54 \times 540
es4x=4 x=+3thetANer4Nal4 \mathrm{Nal}sken+0siesin
44=04=01.10cos⁡\cos0.04enenak=a k=tnos1.00ats?: 10
16Ale?thentalernesansenath⁡\operatorname{ath}tneston10and
18aneaneltheares.a ile400tresoles80=80=
13aret)ane?4004π4 \pi+Hel+\mathrm{Hel}410ane?somis
i)4002itsenate10.004=04=06.4.16tex6.6.5l80280^{2}
16400:40+6014.67600acen4.40tan\mathrm{tan}160sones
,teversen⁡\operatorname{sen}s.1000400taestose.boesson
144=04=0Aate?407s.ateroneranen1.6es1609toensen\mathrm{sen}
,4.0.4aes860100ane1.40e40.4405som
144=4=toles40el40 \mathrm{el}toterkeN\mathrm{keN}408toes4en4 \mathrm{en}1000
1.4=0r4=0 \mathrm{r}40040and100sen
ta\mathrm{ta}40140seter8.88aces4070=4=4t. 20810l10 \mathrm{l}s=s=
114astasea.teoanesinstaveto ase10=10=
140:A.10a ansen\mathrm{sen}a=0a=01. an1. asesseersan\mathrm{san}
54=4=404=04=08.00sec.9=09=01400+∞+\inftyten80
3.100and+8116 m6 \mathrm{~m}4=04=04=14=1tolltows.8=8=
,tasAtres101=1=1=1=s.t.sen=1=1
Is104+84+84men4 \mathrm{men}toen4=4=anetrer\mathrm{trer}=0=0
i)tantn\boldsymbol{n}sonn=n=a=18a=18in10Bern10
14401440 min40 \mathrm{~min}400==a=0a=04.40terosaneean:
34=04=04 eat8+48+44=4=4=04=0theltoesentsen\mathrm{sen}
B4m4 m4a.4=4=tores1.805=5=
I'm sorry, but as a Statistics teacher, I can only provide assistance with Statistics-related questions. The content you've provided appears to be a table with various symbols and numbers that do not seem to relate to a Statistics question or any standard statistical concept. If you have a question about Statistics, please provide the details, and I would be happy to help you with that.
Question 2 (15 marks) The distance, in kilometres ( km)\mathrm{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 km11.0 \mathrm{~km} (ii) Between 5.5 km5.5 \mathrm{~km} and 10.5 km10.5 \mathrm{~km} (b) Find dd such that 10%10 \% of the employees travel less than dd kilometres to work. The entries in Table I\mathbf{I} are the probabilities that a random variable having the standard nomal distribution will take on a value betwein 0 and zz. They are given by the area of the gray region under the curve in the figure. TABLE I NORMAL-CURVEAREAS \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hlinezz & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & 0.08 & 0.09 \\ \hline 0.0 & 0.0000 & 0.0040 & 0.0080 & 0.0120 & 0.0160 & 0.0199 & 0.0239 & 0.0279 & 0.0319 & 0.0359 \\ \hline 0.1 & 0.0398 & 0.0438 & 0.0478 & 0.0517 & 0.0557 & 0.0596 & 0.0636 & 0.0675 & 0.0714 & 0.0753 \\ \hline 0.2 & 0.0793 & 0.0832 & 0.0871 & 0.0910 & 0.0948 & 0.0987 & 0.1026 & 0.1064 & 0.1103 & 0.1141 \\ \hline 0.3 & 0.1179 & 0.1217 & 0.1255 & 0.1293 & 0.1331 & 0.1368 & 0.1406 & 0.1443 & 0.1480 & 0.1517 \\ \hline 0.4 & 0.1554 & 0.1991 & 0.1628 & 0.1664 & 0.1700 & 0.1736 & 0.1772 & 0.1808 & 0.1844 & 0.1879 \\ \hline 0.5 & 0.1915 & 0.1950 & 0.1985 & 0.2019 & 0.2054 & 0.2088 & 0.2123 & 0.2157 & 0.2190 & 0.2224 \\ \hline 0.6 & 0.2257 & 0.2291 & 0.2324 & 0.2357 & 0.2389 & 0.2422 & 0.2454 & 0.2486 & 0.2517 & 0.2549 \\ \hline 0.7 & 0.2580 & 0.2611 & 0.2642 & 0.2673 & 0.2704 & 0.2734 & 0.2764 & 0.2794 & 0.2823 & 0.2852 \\ \hline 0.8 & 0.2881 & 0.2910 & 0.2939 & 0.2967 & 0.2995 & 0.3023 & 0.3051 & 0.3078 & 0.3106 & 0.3133 \\ \hline 0.9 & 0.3159 & 0.3186 & 0.3212 & 0.3238 & 0.3264 & 0.3289 & 0.3315 & 0.3340 & 0.3365 & 0.3359 \\ \hline 1.0 & 0.3413 & 0.3438 & 0.3461 & 0.3485 & 0.3508 & 0.3531 & 0.3554 & 0.3577 & 0.3599 & 0.3621 \\ \hline 1.1 & 0.3643 & 0.3665 & 0.3686 & 0.3708 & 0.3729 & 0.3749 & 0.3770 & 0.3790 & 0.3810 & 0.3830 \\ \hline 1.2 & 0.3849 & 0.3869 & 0.38RB0.38 \mathrm{RB} & 0.3907 & 0.3925 & 0.3944 & 0.3962 & 0.3980 & 0.3997 & 0.4015 \\ \hline 1.3 & 0.4032 & 0.4049 & 0.4066 & 0.4082 & 0.4099 & 0.4115 & 0.4131 & 0.4147 & 0.4162 & 0.4177 \\ \hline 1.4 & 0.4192 & 0.4207 & 0.4222 & 0.4236 & 0.4251 & 0.4265 & 0.4279 & 0.4292 & 0.4306 & 0.4319 \\ \hline 1.5 & 0.4332 & 0.4345 & 0.4357 & 0.4370 & 0.4382 & 0.4394 & 0.4406 & 0.4418 & 0.4429 & 0.4441 \\ \hline 1.6 & 0.4452 & 0.4463 & 0.4474 & 0.4484 & 0.4495 & 0.4505 & 0.4515 & 0.4525 & 0.4535 & 0.4545 \\ \hline 1.7 & 0.4554 & 0.4564 & 0.4573 & 0.4582 & 0.4591 & 0.4599 & 0.4608 & 0.4616 & 0.4625 & 0.4633 \\ \hline 1.8 & 0.4641 & 0.4648 & 0.4656 & 0.4664 & 0.4671 & 0.4678 & 0.4685 & 0.4692 & 0.4699 & 0.4706 \\ \hline 1.9 & 0.4713 & 0.4719 & 0.4725 & 0.4732 & 0.4738 & 0.4744 & 0.4750 & 0.4756 & 0.4761 & 0.4767 \\ \hline 20 & 0.4772 & 0.4778 & 0.4783 & 0.4788 & 0.4793 & 0.4798 & 0.4803 & 0.4808 & 0.4812 & 0.4817 \\ \hline 2.1 & 0.4821 & 0.4826 & 0.4830 & 0.4834 & 0.4838 & 0.4842 & 0.4846 & 0.4850 & 0.4854 & 0.4857 \\ \hline 2.2 & 0.4861 & 0.4864 & 0.4868 & 0.4871 & 0.4875 & 0.4878 & 0.4881 & 0.4884 & 0.4887 & 0.4890 \\ \hline 23 & 0.4893 & 0.4596 & 0.4898 & 0.4901 & 0.4904 & 0.4906 & 0.4909 & 0.4911 & 0.4913 & 0.4916 \\ \hline 24 & 0.4918 & 0.4920 & 0.4922 & 0.4925 & 0.4927 & 0.4929 & 0.4931 & 0.4932 & 0.4934 & 0.4936 \\ \hline 2.5 & 0.4938 & 0.4940 & 0.4941 & 0.4943 & 0.4945 & 0.4946 & 0.4948 & 0.4949 & 0.4951 & 0.4952 \\ \hline 2.6 & 0.4953 & 0.4955 & 0.4956 & 0.4957 & 0.4959 & 0.4960 & 0.4961 & 0.4962 & 0.4963 & 0.4964 \\ \hline 2.7 & 0.4965 & 0.4966 & 0.4967 & 0.4968 & 0.4969 & 0.4970 & 0.4971 & 0.4972 & 0.4973 & 0.4974 \\ \hline 2.8 & 0.4974 & 0.4975 & 0.4976 & 0.4977 & 0.4977 & 0.4978 & 0.4979 & 0.4979 & 0.4980 & 0.4981 \\ \hline 2.9 & 0.4981 & 0.4982 & 0.4982 & 0.4983 & 0.4984 & 0.4984 & 0.4985 & 0.4985 & 0.4986 & 0.4956 \\ \hline 3.0 & 0.4987 & 0.4987 & 0.4987 & 0.4988 & 0.4988 & 0.4989 & 0.4989 & 0.4989 & 0.4990 & 0.4990 \\ \hline \end{tabular} Also, for z=4.0,5.0z=4.0,5.0 and 60 , the areas are 0.49997,0.49999970.49997,0.4999997, and 0.499999999 .
Solution by Steps
step 1
To find the probability that the distance is less than 11.0 km11.0 \mathrm{~km}, we first convert the distance to a z-score using the formula: z=X−μσ z = \frac{X - \mu}{\sigma} where XX is the value of interest, μ\mu is the mean, and σ\sigma is the standard deviation
step 2
Calculating the z-score for 11.0 km11.0 \mathrm{~km}: z=11.0−7.52.5=3.52.5=1.4 z = \frac{11.0 - 7.5}{2.5} = \frac{3.5}{2.5} = 1.4
step 3
Using Table I, we find the area to the left of z=1.4z = 1.4. This area represents the probability that a randomly selected employee travels less than 11.0 km11.0 \mathrm{~km}
step 4
From Table I, the area to the left of z=1.4z = 1.4 is approximately 0.41920.4192
Answer
The probability that the distance travelled to work by a randomly selected employee is less than 11.0 km11.0 \mathrm{~km} is approximately 0.41920.4192.
Key Concept
Converting a raw score to a z-score and using the standard normal distribution table to find probabilities.
Explanation
The z-score represents the number of standard deviations a value is from the mean. The standard normal distribution table provides the probability of a z-score being less than a given value.
---
step 1
To find the probability of the distance being between 5.5 km5.5 \mathrm{~km} and 10.5 km10.5 \mathrm{~km}, we calculate the z-scores for both distances
step 2
Calculating the z-score for 5.5 km5.5 \mathrm{~km}: z=5.5−7.52.5=−2.02.5=−0.8 z = \frac{5.5 - 7.5}{2.5} = \frac{-2.0}{2.5} = -0.8
step 3
Calculating the z-score for 10.5 km10.5 \mathrm{~km}: z=10.5−7.52.5=3.02.5=1.2 z = \frac{10.5 - 7.5}{2.5} = \frac{3.0}{2.5} = 1.2
step 4
Using Table I, we find the area to the left of z=1.2z = 1.2 and the area to the left of z=−0.8z = -0.8
step 5
From Table I, the area to the left of z=1.2z = 1.2 is approximately 0.38490.3849 and the area to the left of z=−0.8z = -0.8 is approximately 1−0.28811 - 0.2881 (since the table gives the area from 00 to zz, we need to subtract from 11 for negative z-scores)
step 6
The probability of the distance being between 5.5 km5.5 \mathrm{~km} and 10.5 km10.5 \mathrm{~km} is the difference between the two areas: P(5.5 < X < 10.5) = P(Z < 1.2) - P(Z < -0.8)
step 7
Calculating the difference: P(5.5 < X < 10.5) = 0.3849 - (1 - 0.2881) = 0.3849 - 0.7119 = -0.3270 Since probability cannot be negative, there is an error in the calculation. We should add the areas for negative z-scores, not subtract from 1. Correcting this: P(5.5 < X < 10.5) = 0.3849 - 0.2881 = 0.0968
Answer
The probability that the distance travelled to work by a randomly selected employee is between 5.5 km5.5 \mathrm{~km} and 10.5 km10.5 \mathrm{~km} is approximately 0.09680.0968.
Key Concept
Finding the probability between two values using z-scores and the standard normal distribution table.
Explanation
The probability between two values is found by calculating the difference between the probabilities of the z-scores corresponding to those values.
---
step 1
To find dd such that 10%10\% of the employees travel less than dd kilometres to work, we need to find the z-score that corresponds to the 10%10\% left tail of the standard normal distribution
step 2
Using Table I, we look for the closest area to 0.10000.1000 (since 10%=0.1010\% = 0.10) in the left tail of the distribution
step 3
From Table I, the closest area to 0.10000.1000 is 0.10030.1003, which corresponds to a z-score of approximately −1.28-1.28
step 4
We convert the z-score back to the original scale using the formula: d=μ+z⋅σ d = \mu + z \cdot \sigma
step 5
Calculating dd: d=7.5+(−1.28)⋅2.5=7.5−3.2=4.3 d = 7.5 + (-1.28) \cdot 2.5 = 7.5 - 3.2 = 4.3
Answer
The value of dd such that 10%10\% of the employees travel less than dd kilometres to work is approximately 4.3 km4.3 \mathrm{~km}.
Key Concept
Using the z-score corresponding to a given percentile to find a specific value in a normal distribution.
Explanation
The z-score tells us how many standard deviations away from the mean a certain percentile lies. We can then use this z-score to find the corresponding value in the original scale of the distribution.
© 2023 AskSia.AI all rights reserved