Probability And Statistics Applications

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Probability and Statistics Applications1)In a particular suburb, 30% of the households have installed electronic securitysystems.a)If a household is chosen at random from this suburb, what is the probabilitythat this household has not installed anelectronic security system?b)If two households are chosen at random from this suburb, what is theprobability that neither has installed an electronic security system?Answer:a)If 30% of households have installed an electronic security system,then the probability that a randomly chosen householdhas notinstalled anelectronic security system is:P(Nosecuritysystem)=1−P(Hassecuritysystem)=1−0.30=0.70P(\text{No securitysystem}) = 1-P(\text{Has security system}) = 1-0.30 = 0.70So, the probability that the household has not installed an electronic security systemis0.70.b)If two households are chosen at random, and both are independent of each other,the probability thatneitherhousehold has installed an electronic security system is:P(Neitherhassecuritysystem)=P(Nosecuritysystemfor1sthousehold)×P(Nosecuritysystemfor2ndhousehold)P(\text{Neither has security system}) = P(\text{Nosecurity system for 1st household})\times P(\text{No security system for 2ndhousehold}) =0.70×0.70=0.49= 0.70\times 0.70 = 0.49So, the probability that neither household has installed an electronic security systemis0.49.5)In a study designed to gauge married women’s participation in the workplacetoday, the data provided in the fileP04_05.xlsxwere obtained from a sample of 750randomly selected married women. Consider a woman selected at random from thissample in answering the following questions.a)What is the probability that this randomly selected woman has a job outsidethe home?b)What is the probability that this randomly selected woman has at least onechild?c)What is the probability that this randomly selected woman has a full-time joband no more than one child?d)What is the probability that this randomly selected woman has a part-time jobor at least one child, but not both?

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**HINT**(n=no job, P=part time, F=full time)Answer:To answer these questions, we would typically need to refer to thedataset in the fileP04_05.xlsxto get the necessary data. However, since thefile is not available here, I will provide you with a general method for solvingeach part of the problem:a) Probability that the selected woman has a job outside the homeTo find this probability, you need to divide the number of women who have ajob outside the home by the total sample size (750). The formula is:P(Joboutsidethehome)=Numberofwomenwithajoboutsidethehome750P(\text{Job outside the home}) =\frac{\text{Number of women with a joboutside the home}}{750}Refer to the data in the file for the exact count of women with jobs.b) Probability that the selected woman has at least one childTo find this probability, divide the number of women who have at least onechild by the total sample size (750). The formula is:P(Atleastonechild)=Numberofwomenwithatleastonechild750P(\text{Atleast one child}) =\frac{\text{Number of women with at least one child}}{750}Again, you’ll need the specific count from the file for this calculation.c) Probability that the selected woman has a full-time job and no morethan one childThis is a joint probability problem. You need to find the number of women whohave a full-time jobandhave no more than one child. The formula is:P(Full-timejoband≤1child)=Numberofwomenwithfull-timejobsandnomorethanonechild750P(\text{Full-time job and }\leq 1\text{ child}) =\frac{\text{Number of women with full-time jobs and no morethan one child}}{750}This will require checking the counts for full-time job and child status in thedataset.d) Probability that the selected woman has a part-time job or at leastone child, but not bothThis is a more complex question because we need to calculate the probabilityof the union of two events,part-time job or at least one child, but excludingthe overlap (i.e., women who have both). The formula is:P(Part-timeoratleastonechild,butnotboth)=P(Part-time)+P(Atleastonechild)−2×P(Part-timeandatleastonechild)P(\text{Part-time or at least one child, but not both}) = P(\text{Part-time}) + P(\text{At leastone child})-2\times P(\text{Part-time and at least one child})

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Where:•P(Part-time)=Numberofwomenwithpart-timejobs750P(\text{Part-time}) =\frac{\text{Number of women with part-time jobs}}{750}•P(Atleastonechild)=Numberofwomenwithatleastonechild750P(\text{At leastone child}) =\frac{\text{Number of women with at least one child}}{750}•P(Part-timeandatleastonechild)=Numberofwomenwithpart-timejobsandatleastonechild750P(\text{Part-time and at least one child}) =\frac{\text{Number of women with part-time jobs and at least one child}}{750}Again, use the data from the file to get these counts.Summary of the Steps:1.For each part: Extract the necessary counts from the dataset.2.Apply the formulasprovided to calculate the probabilities.3.If you have access to a program (Excel, R, Python, etc.), you can quicklycompute thesevalues using the provided counts.10)A fair coin (i.e., heads and tails are equally likely) is tossed three times. Let X bethe number of heads observed in three tosses of this fair coin.a)Find the probability distribution of X.b) Find the probability that two or fewer heads are observed in three tosses.c) Find the probability that at least one head is observed in three tosses.d) Find the expected value of X.e) Find the standard deviation of X.Answer:Problem: A fair coin is tossed threetimes, and XX representsthe number of heads observed in those tosses.Let's break down each part of the problem.a) Find the probability distribution of XXThe random variable XX represents the number of heads observed in 3tosses of a fair coin. The possible values of XX are 0,1,2,0, 1, 2, and 33,since there can be anywhere from 0 to 3 heads in 3 tosses.To find the probability distribution, we first need to calculate the probabilitiesfor each value of XX.Each coin toss has 2 outcomes: heads (H) or tails (T). Since there are 3tosses, the total number of possible outcomes is 23=82^3 = 8.The sample space SS consists of all possible outcomes:S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}S =\{HHH, HHT, HTH, HTT,THH, THT, TTH, TTT\}Now, count how many of these outcomes correspond to each possible valueof XX:•X=0X = 0 (no heads): There is 1 outcome: TTTTTT.

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•X=1X = 1 (one head): There are 3 outcomes: HTT,THT,TTHHTT, THT, TTH.•X=2X = 2 (two heads): There are 3 outcomes: HHT,HTH,THHHHT, HTH, THH.•X=3X = 3(three heads): There is 1 outcome: HHHHHH.Now, calculate the probabilities:•P(X=0)=18P(X = 0) =\frac{1}{8}•P(X=1)=38P(X = 1) =\frac{3}{8}•P(X=2)=38P(X = 2) =\frac{3}{8}•P(X=3)=18P(X = 3) =\frac{1}{8}So, the probability distribution of XX is:P(X=0)=18,P(X=1)=38,P(X=2)=38,P(X=3)=18P(X = 0) =\frac{1}{8},\quad P(X= 1) =\frac{3}{8},\quad P(X = 2) =\frac{3}{8},\quad P(X = 3) =\frac{1}{8}b) Find the probability that two or fewer heads are observed in threetosses.We need to find P(X≤2)P(X\leq 2), which is the probability of observing 0, 1,or 2 heads. Using the probability distribution from part (a):P(X≤2)=P(X=0)+P(X=1)+P(X=2)P(X\leq 2) = P(X = 0) + P(X = 1) + P(X = 2)P(X≤2)=18+38+38=78P(X\leq 2) =\frac{1}{8} +\frac{3}{8} +\frac{3}{8} =\frac{7}{8}So, the probability of observing two or fewer heads is 78\frac{7}{8}.c) Find the probability that at least one head is observed in three tosses.We need to find P(X≥1)P(X\geq 1), which is the probability of observing atleast one head. This is the complement of the probability of observing zeroheads.P(X≥1)=1−P(X=0)P(X\geq 1) = 1-P(X = 0) P(X≥1)=1−18=78P(X\geq 1) = 1-\frac{1}{8} =\frac{7}{8}So, the probability of observing at least one head is 78\frac{7}{8}.d) Find the expected value of XX.The expected value E(X)E(X) is calculated as the sum of each value of XXmultiplied by its corresponding probability:E(X)=(0×P(X=0))+(1×P(X=1))+(2×P(X=2))+(3×P(X=3))E(X) = (0\times P(X =0)) + (1\times P(X = 1)) + (2\times P(X = 2)) + (3\times P(X = 3))E(X)=(0×18)+(1×38)+(2×38)+(3×18)E(X) = (0\times\frac{1}{8}) + (1\times\frac{3}{8}) + (2\times\frac{3}{8}) + (3\times\frac{1}{8})E(X)=0+38+68+38=128=1.5E(X) = 0 +\frac{3}{8} +\frac{6}{8} +\frac{3}{8} =\frac{12}{8} = 1.5So, the expected value of XX is 1.51.5.e) Find the standard deviation of XX.The standard deviation σ\sigma is the square root of the variance. To find thevariance, we first calculate E(X2)E(X^2) and then use the formula:

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σ2=E(X2)−(E(X))2\sigma^2 = E(X^2)-(E(X))^2First, calculate E(X2)E(X^2), which is the sum of each value of X2X^2multiplied by its corresponding probability:E(X2)=(02×P(X=0))+(12×P(X=1))+(22×P(X=2))+(32×P(X=3))E(X^2) = (0^2\times P(X = 0)) + (1^2\times P(X = 1)) + (2^2\times P(X = 2)) + (3^2\timesP(X = 3)) E(X2)=(0×18)+(1×38)+(4×38)+(9×18)E(X^2) = (0\times\frac{1}{8})+ (1\times\frac{3}{8}) + (4\times\frac{3}{8}) + (9\times\frac{1}{8})E(X2)=0+38+128+98=248=3E(X^2) = 0 +\frac{3}{8} +\frac{12}{8} +\frac{9}{8} =\frac{24}{8} = 3Now, calculate the variance:σ2=E(X2)−(E(X))2=3−(1.5)2=3−2.25=0.75\sigma^2 = E(X^2)-(E(X))^2 = 3-(1.5)^2 = 3-2.25 = 0.75Finally, the standard deviation is the square root of the variance:σ=0.75≈0.866\sigma =\sqrt{0.75}\approx 0.866So, thestandard deviation of XX is approximately 0.8660.866.Summary of Answers:•a)Probability distribution of XX:P(X=0)=18,P(X=1)=38,P(X=2)=38,P(X=3)=18P(X = 0) =\frac{1}{8},\quad P(X= 1) =\frac{3}{8},\quad P(X = 2) =\frac{3}{8},\quad P(X = 3) =\frac{1}{8}•b)Probability of two or fewer heads: 78\frac{7}{8}•c)Probability of at least one head: 78\frac{7}{8}•d)Expected value of XX: 1.51.5•e)Standard deviation of XX: ≈0.866\approx 0.86611)Consider a random variable with the following probability distribution: P(X=0) =0.1, P(X =1) = 0.2, P(X = 2) = 0.3, P(X = 3) = 0.3, and P(X = 4) = 0.1a)Find P(X<2)b) Find P(1< X<3)c) Find P(X > 0)d) Find P(X > 3/X > 2)e) Find the expected value of X.f) Find the standard deviation of X.Answer:Let's work through each part of the problem, step by step.Given Probability Distribution:P(X=0)=0.1,P(X=1)=0.2,P(X=2)=0.3,P(X=3)=0.3,P(X=4)=0.1P(X = 0) = 0.1,\quad P(X = 1) = 0.2,\quad P(X = 2) = 0.3,\quad P(X = 3) = 0.3,\quad P(X =4) = 0.1

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