Lecture 23. STAT 225 Introduction to Probability Models April 4, Whitney Huang Purdue University. Normal approximation to Binomial

Transcription

1 Lecture 23 STAT 225 Introduction to Probability Models April 4, 2014 approximation Whitney Huang Purdue University 23.1

2 Agenda 1 approximation 2 approximation 23.2

3 Characteristics of the random variable: Let X be a r.v. The support for X: (, ) approximation 23.3

4 Characteristics of the random variable: Let X be a r.v. The support for X: (, ) Its parameter(s) and definition(s): µ : mean and σ 2 : variance approximation 23.3

5 Characteristics of the random variable: Let X be a r.v. The support for X: (, ) Its parameter(s) and definition(s): µ : mean and σ 2 : variance 1 The probability density function (pdf): 2πσ e (x µ)2 2σ 2 < x < for approximation 23.3

6 Characteristics of the random variable: Let X be a r.v. The support for X: (, ) Its parameter(s) and definition(s): µ : mean and σ 2 : variance 1 The probability density function (pdf): 2πσ e (x µ)2 2σ 2 < x < The cumulative distribution function (cdf): No explicit form, look at the value Φ( x µ σ ) for < x < from standard normal table for approximation 23.3

7 Characteristics of the random variable: Let X be a r.v. The support for X: (, ) Its parameter(s) and definition(s): µ : mean and σ 2 : variance 1 The probability density function (pdf): 2πσ e (x µ)2 2σ 2 < x < The cumulative distribution function (cdf): No explicit form, look at the value Φ( x µ σ ) for < x < from standard normal table The expected value: E[X] = µ for approximation 23.3

8 Characteristics of the random variable: Let X be a r.v. The support for X: (, ) Its parameter(s) and definition(s): µ : mean and σ 2 : variance 1 The probability density function (pdf): 2πσ e (x µ)2 2σ 2 < x < The cumulative distribution function (cdf): No explicit form, look at the value Φ( x µ σ ) for < x < from standard normal table The expected value: E[X] = µ The variance: Var(X) = σ 2 for approximation 23.3

9 Standard Z N(µ = 0, σ 2 = 1) random variable X with mean µ and standard deviation σ can convert to standard normal Z by the following : Z = X µ σ approximation 23.4

10 Standard Z N(µ = 0, σ 2 = 1) random variable X with mean µ and standard deviation σ can convert to standard normal Z by the following : Z = X µ σ The cdf of the standard normal, denoted by Φ(z), can be found from the standard normal table approximation 23.4

11 Standard Z N(µ = 0, σ 2 = 1) random variable X with mean µ and standard deviation σ can convert to standard normal Z by the following : Z = X µ σ The cdf of the standard normal, denoted by Φ(z), can be found from the standard normal table The probability P(a X b) where X N(µ, σ 2 ) can be compute approximation P(a X b) = P( a µ σ = Φ( b µ σ ) Φ(a µ σ ) Z b µ σ ) 23.4

12 Properties of Φ Φ(0) =.50 Mean and Median (50 th percentile) for standard normal are both 0 approximation 23.5

13 Properties of Φ Φ(0) =.50 Mean and Median (50 th percentile) for standard normal are both 0 Φ( z) = 1 Φ(z) approximation 23.5

14 Properties of Φ Φ(0) =.50 Mean and Median (50 th percentile) for standard normal are both 0 Φ( z) = 1 Φ(z) P(Z > z) = 1 Φ(z) = Φ( z) approximation 23.5

15 Sums of Random Variables If X i 1 i n are independent normal random variables with mean µ i are variance σi 2, respectively. Let S n = n i=1 X i then S n N( n i=1 µ i, n i=1 σ2 i ) approximation 23.6

16 Sums of Random Variables If X i 1 i n are independent normal random variables with mean µ i are variance σi 2, respectively. Let S n = n i=1 X i then S n N( n i=1 µ i, n i=1 σ2 i ) This can be applied for any integer n approximation 23.6

17 The Empirical Rules The Empirical Rules are a way to approximate certain probabilities for the as the following table: Interval Percentage with interval µ ± σ 68% µ ± 2σ 95% µ ± 3σ 99.7% approximation 23.7

18 approximation of We can use a to approximate a if n is large and p is close to.5 approximation 23.8

19 approximation of We can use a to approximate a if n is large and p is close to.5 Rule of thumb for this be valid (in this class) is np > 5 and n(1 p) > 5 approximation 23.8

20 approximation of We can use a to approximate a if n is large and p is close to.5 Rule of thumb for this be valid (in this class) is np > 5 and n(1 p) > 5 If X (n, p) with np > 5 and n(1 p) > 5 then we can use X N(µ = np, σ 2 = np(1 p)) to approximate X approximation 23.8

21 approximation of We can use a to approximate a if n is large and p is close to.5 Rule of thumb for this be valid (in this class) is np > 5 and n(1 p) > 5 If X (n, p) with np > 5 and n(1 p) > 5 then we can use X N(µ = np, σ 2 = np(1 p)) to approximate X Notice that is a discrete distribution but normal is a continuous distribution so that P(X = x) = 0 x approximation 23.8

22 approximation of We can use a to approximate a if n is large and p is close to.5 Rule of thumb for this be valid (in this class) is np > 5 and n(1 p) > 5 If X (n, p) with np > 5 and n(1 p) > 5 then we can use X N(µ = np, σ 2 = np(1 p)) to approximate X Notice that is a discrete distribution but normal is a continuous distribution so that P(X = x) = 0 x By using continuity correction we use P(x 0.5 X x + 0.5) to approximate P(X = x) approximation 23.8

23 Example 57 For this entire problem, please use the empirical rules. The number of pairs of shoes in an adult female s closet is with a mean of 58 and a standard deviation of 5 1 What interval contains the middle 68% of the distribution? 2 Find the value such that 2.5% are lower than that value approximation 3 What is the probability an adult female s closet has between 48 and 63 pairs of shoes? 23.9

24 Example 57 cont d Solution. 1 (58 5, ) = (53, 63) = % + 95% 68% 2 = 81.5% approximation 23.10

25 Example 58 Suppose a class has 400 students (to begin with), that each student drops independently of any other student with a probability of.07. Let X be the number of students that finish this course 1 Find the probability that X is between 370 and 373 inclusive 2 Is an approximation appropriate for the number of students that finish the course? 3 If so, what is this distribution and what are the value(s) of its parameter(s)? 4 Find the probability that is between 370 and 373 inclusive by using the approximation (if an approximation appropriate) approximation 23.11

26 Example 58 cont d Solution. X (n = 400,.93) 1 P(370 X 373) =.3010 approximation 23.12

27 Example 58 cont d Solution. X (n = 400,.93) 1 P(370 X 373) = Yes, since np = 372 > 5 and n(1 p) = 28 > 5. The normal approximation for binomial is appropriate approximation 23.12

28 Example 58 cont d Solution. X (n = 400,.93) 1 P(370 X 373) = Yes, since np = 372 > 5 and n(1 p) = 28 > 5. The normal approximation for binomial is appropriate 3 X N(µ = 372, σ 2 = 400(.93)(.07) = 26.04) approximation 23.12

29 Example 58 cont d Solution. X (n = 400,.93) 1 P(370 X 373) = Yes, since np = 372 > 5 and n(1 p) = 28 > 5. The normal approximation for binomial is appropriate 3 X N(µ = 372, σ 2 = 400(.93)(.07) = 26.04) approximation 4 P(369.5 X 373.5) = P(.49 Z.29) = =