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Probability And Statistics Applications

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    Probability and Statistics Applications 1) In a particular suburb, 30% of the households have installed electronic security systems. a) If a household is chosen at random from this suburb, what is the probability that this household has not installed an electronic security system? b) If two households are chosen at random from this suburb, what is the probability 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 household has not installed an electronic security system is: P(No security system)=1−P(Has security system)=1−0.30=0.70P( \ text{No security system}) = 1 - P( \ text{Has security system}) = 1 - 0.30 = 0.70 So, the probability that the household has not installed an electronic security system is 0.70 . b) If two households are chosen at random, and both are independent of each other, the probability that neither household has installed an electronic security system is: P(Neither has security system)=P(No security system for 1st household)×P(No secu rity system for 2nd household)P( \ text{Neither has security system}) = P( \ text{No security system for 1st household}) \ times P( \ text{No security system for 2nd household}) =0.70×0.70=0.49= 0.70 \ times 0.70 = 0.49 So, the probability that neither household has installed an electronic security system is 0.49 . 5) In a study designed to gauge married women’s participation in the workplace today, the data provided in the file P0 4_05.xlsx were obtained from a sample of 750 randomly selected married women. Consider a woman selected at random from this sample in answering the following questions. a) What is the probability that this randomly selected woman has a job outside the home? b) What is the probability that this randomly selected woman has at least one child? c) What is the probability that this randomly selected woman has a full - time job and no more than one child? d) What is the probability that this randomly selected woman has a part - time job or 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 the dataset in the file P04_05.xlsx to get the necessary data. However, since the file is not available here, I will provide you with a general method for solving each part of the problem: a) Probability that the selected woman has a job outside the home To find this probability, you need to divide the number of women who have a job outside the home by the total sample size (750). The formula is: P(Job outside the home)=Number of women with a job outside the home750P ( \ text{Job outside the home}) = \ frac{ \ text{Number of women with a job outside 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 child To find this probability, divide the number of women who have at least one child by the total sample size (750). The formula is: P(At least one child)=Number of women with at least one child750P( \ text{At least 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 more than one child This is a joint probability problem. You need to find the number of women who have a full - time job and have no more than one child. The formula is: P(Full - time job and ≤1 child)=Number of women with full - time jobs and no more than one child750P( \ text{Full - time job and } \ leq 1 \ text{ child}) = \ frac{ \ text{Number of women with full - time jobs and no more than one child}}{750} This will require checking the counts for full - time job and child status in the dataset. d) Probability that the selected woman has a part - time job or at least one child, but not both This is a more complex question because we need to calculate the probability of the union of two events, part - time job or at least one child , but excluding the overlap (i.e., women who have both). The formula is: P(Part - time or at least one child, but not both)=P(Part - time)+P(At least one child)−2×P(Part - time and at least one child)P( \ text{Part - time or at least one child, but not both}) = P( \ text{Part - time}) + P( \ text{At least one child}) - 2 \ times P( \ text{Part - time and at least one child})

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    Where: • P(Part - time)=Number of women with part - time jobs750P( \ text{Part - time}) = \ frac{ \ text{Number of women with part - time jobs}}{750} • P(At least one child)=Number of women with at least one child750P( \ text{At least one child}) = \ frac{ \ text{Number of women with at least one child}}{750} • P(Part - time and at least one child)=Number of women with part - time jobs and at least one child750P( \ 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 formulas provided to calculate the probabilities. 3. If you have access to a program (Excel, R, Python, etc.), you can quickly compute these values using the provided counts. 10) A fair coin (i.e., heads and tails are equally likely) is tossed three times. Let X be the 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 three times, and XX represents the number of heads observed in those tosses. Let's break down each part of the problem. a) Find the probability distribution of XX The random variable XX represents the number of heads observed in 3 tosses 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 probabilities for each value of XX. Each coin toss has 2 outcomes: heads (H) or tails (T). Since there are 3 tosses, 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 value of 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 three tosses. 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 at least one head. This is the complement of the probability of observing zero heads. 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 XX multiplied 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} = 1.5 So, 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 the variance, 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))^2 First, calculate E(X2)E(X^2), which is the sum of each value of X2X^2 multiplied 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 \ times P(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} = 3 Now, 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 = 0.75 Finally, the standard deviation is the square root of the variance: σ=0.75≈0.866 \ sigma = \ sqrt{0.75} \ approx 0.866 So, the standard 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 0.866 11) Consider a random variable with the following probability distribut ion: P(X=0) = 0.1, P(X =1) = 0.2, P(X = 2) = 0.3, P(X = 3) = 0.3, and P(X = 4) = 0.1 a) 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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