Chapter 4 part2- Random Variables

Chapter 4 part2- Random Variables

• 1.
  INTRODUCTION TO STATISTICS &PROBABILITY Chapter 4: Probability: The Study of Randomness (Part 2) Dr. Nahid Sultana 1
• 2.
  Chapter 4 Probability: TheStudy of Randomness 4.1 Randomness 4.2 Probability Models 4.3 Random Variables 4.4 Means and Variances of Random Variables 4.5 General Probability Rules* 2
• 3.
  4.3 Random Variables 3 Random Variable  Discrete Random Variables  Continuous Random Variables  Normal Distributions as Probability Distributions
• 4.
  4 Random Variables 4  Aprobability model: sample space S and probability for each outcome.  A numerical variable that describes the outcomes of a chance process is called a random variable.  The probability model for a random variable is its probability distribution. The probability distribution of a random variable gives its possible values and their probabilities. Example: Consider tossing a fair coin 3 times. Define X = the number of heads obtained. X = 0: TTT X = 1: HTT THT TTH X = 2: HHT HTH THH X = 3: HHH Value 0 1 2 3 Probability 1/8 3/8 3/8 1/8
• 5.
  5 Discrete Random Variable Twomain types of random variables: discrete and continuous. A discrete random variable X takes a fixed set of possible values with gaps between. The probability distribution of a discrete random variable X lists the values xi and their probabilities pi: The probabilities pi must satisfy two requirements: 1. Every probability pi is a number between 0 and 1. 2. The sum of the probabilities is 1.
• 6.
  6 Discrete Random Variable(Cont…) Example: Consider tossing a fair coin 3 times. Define X = the number of heads obtained. X = 0: TTT X = 1: HTT THT TTH X = 2: HHT HTH THH X = 3: HHHValue 0 1 2 3 Probability 1/8 3/8 3/8 1/8 Q1: What is the probability of tossing at least two heads? Ans: P(X ≥ 2 ) = P(X=2) + P(X=3) = 3/8 + 1/8 = 1/2 Q2: What is the probability of tossing fewer than three heads? Ans: P(X < 3 ) = P(X=0) +P(X=1) + P(X=2) = 1/8 + 3/8 + 3/8 = 7/8 Or P(X < 3 ) = 1 – P(X = 3) = 1 – 1/8 = 7/8
• 7.
  7 Discrete Random Variable(Cont…) Example: North Carolina State University posts the grade distributions for its courses online. Students in one section of English210 in the spring 2006 semester received 31% A’s, 40% B’s, 20% C’s, 4% D’s, and 5% F’s. The student’s grade on a four-point scale (with A = 4) is a random variable X. The value of X changes when we repeatedly choose students at random , but it is always one of 0, 1, 2, 3, or 4. Here is the distribution of X: Q1: What is the probability that the student got a B or better? Ans: P(X ≥ 3 ) = P(X=3) + P(X=4) = 0.40 + 0.31 = 0.71 Q2: Suppose that a grade of D or F in English210 will not count as satisfying a requirement for a major in linguistics. What is the probability that a randomly selected student will not satisfy this requirement? Ans: P(X ≤ 1 ) = 1 - P( X >1) = 1 – ( P(X=2) + P(X=3) + P(X=4) ) = 1- 0.91 = 0.09
• 8.
  8 Continuous Random Variable Acontinuous random variable Y takes on all values in an interval of numbers. Ex: Suppose we want to choose a number at random between 0 and 1. -----There is infinitely many number between 0 and 1. How do we assign probabilities to events in an infinite sample space?  The probability distribution of Y is described by a density curve.  The probability of any event is the area under the density curve and above the values of Y that make up the event.
• 9.
  9  A discreterandom variable X has a finite number of possible values. The probability model of a discrete random variable X assigns a probability between 0 and 1 to each possible value of X.  A continuous random variable Y has infinitely many possible values. The probability of a single event (ex: X=k) is meaningless for a continuous random variable. Only intervals can have a non-zero probability; represented by the area under the density curve for that interval .  Discrete random variables commonly arise from situations that involve counting something.  Situations that involve measuring something often result in a continuous random variable. Continuous Random Variable (Cont…)
• 10.
  10 Continuous Probability Models Example:This is a uniform density curve for the variable X. Find the probability that X falls between 0.3 and 0.7. Ans: P(0.3 ≤ X ≤ 0.7) = (0.7- 0.3) * 1 = 0.4 Uniform Distribution
• 11.
  11 Continuous Probability Models(Cont…) Example: Find the probability of getting a random number that is less than or equal to 0.5 OR greater than 0.8. P(X ≤ 0.5 or X > 0.8) = P(X ≤ 0.5) + P(X > 0.8) = 0.5 + 0.2 = 0.7 Uniform Distribution
• 12.
  12 Continuous Probability Models(Cont…) General Form: The probability of the event A is the shaded area under the density curve. The total area under any density curve is 1.
• 13.
  13 Normal Probability Model Theprobability distribution of many random variables is a normal distribution. Example: Probability distribution of Women’s height. Here, since we chose a woman randomly, her height, X, is a random variable. To calculate probabilities with the normal distribution, we standardize the random variable (z score) and use the Table A.
• 14.
  14 Normal Probability Model(Cont…) Reminder: standardizing N(µ,σ) We standardize normal data by calculating z-score so that any normal curve can be transformed into the standard Normal curve N(0,1). σ µ)( − = x z
• 15.
  15 Normal Probability Model (Cont…) Women’sheights are normally distributed with µ = 64.5 and σ = 2.5 in. The z-scores for 68, And for x = 70", 4.1 5.2 )5.6468( = − =z z = (70−64.5) 2.5 = 2.2 The area under the curve for the interval [68”,70”] is 0.9861-0.9192=0.0669. Thus the probability that a randomly chosen woman falls into this range is 6.69%. i.e. P(68 ≤ X ≤ 70)= 6.69%. What is the probability, if we pick one woman at random, that her height will be between 68 and 70 inches i.e. P(68 ≤ X ≤ 70)? Here because the woman is selected at random, X is a random variable.