Statistics AI Solver
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Sample Questions
Expert help when you need it- Q1: Let the random variable X equal to the number of days that it takes a high-risk driver to have an accident. Assume that X has an exponential distribution. If P(X<100)=0.25, compute a) P(X>200) b) P(X> 150 >50) c) E(X)AnswerSolve for using P(X<100)=0.25 then find P(X>200) Cumulative distribution function. P(X
- Q2:A bus arrives every 10 minutes at a bus stop. Passengers arrive at the bus stop at random times. a) What is the probability that the individual waits more than 6 minutes? b) Given that an individual has been waiting for 3 minutes, what is the probability that the individual will get on the bus in the next 5 minutes? c) Say an individual takes this bus everyday in a month, what’s the probability that the 20th day of the month is the 3rd time that they have to wait for more than 8 minutes?Answera) Since a bus arrival follow a uniform distribution between 0-10 minutes. P(wait>6)=10-610=410=0.4 b) If the person has waited 3 mons, there are 7 min left before the next bus arrives. Probability of getting the next bus arrives in 5 minutes =5/7 c) P(wait>8)=10-810=0.2 We must find probability that the 20th day is the 3rd time the experience a wait longer than 8 min P(X=3)= (203)*(0.2)^3 (0.8)^17 Binomial(20,0.2) 203=20!/3!(20-3)!=20*19*18/3*2*1=1140 P(X=3)= 1140*(0.2)^3 (0.8)^17=0.353
- Q3:Consider a set S = {2, 4, 6, 8, x, y} with distinct elements. If x and y are both prime numbers and 0 < x < 40 and 0 < y < 40, which of the following MUST be true? I. The maximum possible range of the set is greater than 33. II. The median can never be an even number. III. If y = 37, the average of the set will be greater than the medianAnswerGiven S = {2, 4, 6, 8, x, y} x and y are prime numbers (0 < x < 40) and (0 < y < 40). -The set S contains distinct elements . We'll analyze each statement to determine its validity. --- Statement I: The maximum possible range of the set is greater than 33 The range of a set is defined as: {Range} = {Maximum value} - {Minimum value}. The smallest values in (S) are fixed as (2, 4, 6, 8). To maximize the range: - Set (x) and (y) to the largest primes less than 40. These are (37) and (31). - Place (x = 31) and (y = 37) (or vice versa). The largest value in \(S\) will be \(37\), and the smallest value will be (2). Thus, the range is: {Range} = 37 - 2 = 35. Since (35 > 33), Statement I is true. --- Statement II: The median can never be an even number. The median is the middle value (or the average of the two middle values) when the elements of the set are arranged in ascending order. The set S always contains the even numbers (2, 4, 6, 8) and two primes (x, y). Since there are 6 elements in total: - The median is the average of the 3rd and 4th smallest numbers in the sorted set. Case Analysis: 1. If (x) and (y) are smaller primes, like (3) and (5): The sorted set becomes {2, 3, 4, 5, 6, 8}. The median is: {Median} = {4 + 5}/{2} = 4.5 2. If (x) and (y) are larger primes, like (31) and (37): The sorted set becomes {2, 4, 6, 8, 31, 37}. The median is: {Median} = {6 + 8}/{2} = 7. In either case, the median could be an even number (if the two middle numbers are both even, as in the case {2, 4, 6, 8, x, y}. Statement II is false. --- Statement III: If (y = 37), the average of the set will be greater than the median. If (y = 37), the set becomes {2, 4, 6, 8, x, 37}(assuming \(x\) is any prime number less than 40 and distinct from (37)). Step 1: Calculate the Average The average is given by: {Average} = {Sum of all elements}/{Number of elements} Let the sum of \(S\) be \(2 + 4 + 6 + 8 + x + 37\): {Average} = {57 + x}/{6} : Analyze the Median The median is the average of the 3rd and 4th smallest numbers in the sorted set. If (x) is small (e.g., (x = 3)), the sorted set is {2, 3, 4, 6, 8, 37}, and the median is: {Median} = {4 + 6}/{2} = 5. If x is large (e.g., (x = 31)), the sorted set is {2, 4, 6, 8, 31, 37}, and the median is: {Median} = {6 + 8}/{2} = 7. Step 3: Compare the Average and Median 1. For (x = 3): {Average} ={57 + 3}/{6} = 10. Since (10 > 5), the average is greater than the median. 2. For (x = 31): {Average} = {57 + 31}/{6} = 14.67. Since (14.67 > 7), the average is greater than the median. Thus, for all valid (x), the average will always be greater than the median when (y = 37). Statement III is true. Statement I: True Statement II: False Statement III: True
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