The binomial distribution p314

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1 The binomial distribution p314 Example: A biased coin (P(H) = p = 0.6) ) is tossed 5 times. Let X be the number of H s. Fine P(X = 2). This X is a binomial r. v. The binomial setting p There are a fixed number n of observations. 2. The n observations are independent. 3. Each observation falls into one of just two categories (successes and failures) 4. The probability of a success (call it p) is the same for each observation. 1

2 Probability function of the binomial dist. If X has a B(n, p), n P( X x) px(1 p) nx for n0,1,, n x Mean and Variance of a binomial r. v. p320 If X has a B(n, p) np and 2 np(1 p) X X 2

3 Probability Mass Binomial Distribution: Binomial trials=5, Probability of success= Number of Successes 3

4 Probability Mass Probability Mass Binomial Distribution: Binomial trials=20, Probability of success= Number of Successes Binomial Distribution: Binomial trials=30, Probability of success= Number of Successes 4

5 Probability Mass Binomial Distribution: Binomial trials=500, Probability of success= Number of Successes 5

6 Sampling distribution of a count p316 When the population is much larger than the sample (at least 20 times larger), the count X of successes in a SRS of size n has approximately B(n, p) where p is the population proportion of successes. 6

7 S5.1 You are planning a sample survey of small businesses in your area. You will choose an SRS of businesses listed in the telephone book's Yellow Pages. Experience shows that only about half the businesses you contact will respond. (a) If you contact 150 businesses, it is reasonable to use the B(150; 0:5) distribution for the number X who respond. Explain why. (b) What is the expected number (the mean) who will respond and what is its std dev.? 7

8 Ex. The probability that a certain machine will produce a defective item is 1/4. If a random sample of 6 items is taken from the output of this machine, what is the probability that there will be 5 or more defectives in the sample? 8

9 Ex There are 20 multiple-choice questions on an exam, each having responses a, b, c, and d. Each question is worth 5 points. And only one response per question is correct. Suppose that a student guesses the answer to question and her guesses from question to question are independent. It the student needs at least 40 points to pass the test. What is the probability that the student will pass the test? Ans. X~B(20, 0.25). P(X>=8) = , adding the entries 8 through 20 in the appropriate of table C What is the expected (mean) score for this student. Ans. 20 x 0.25 = 5 and expected score =5 x 5 = 25 9

10 Sample Proportions p321 Sample proportion ( ˆp) = X n Mean and std dev of a sample proportion of successes in a SRS of size n. p and p(1 p) pˆ pˆ n 10

11 Normal approximation for counts and proportions P323 Draw a SRS of size n from a large population having population p of success. Let X be the count of success in the sample and pˆ X / n the sample proportion of successes. When n is large, the sampling distributions of these statistics are approximately normal: X is approx. N( np, np(1 p)) ˆp is approx. N p, p(1 p) n As a rule of thumb, we will use this approximation for values of n and p that satisfy np 10 and n(1 p) 10. S5.1 You are planning a sample survey of small businesses in your 11

12 area. You will choose an SRS of businesses listed in the telephone book's Yellow Pages. Experience shows that only about half the businesses you contact will respond. (a) If you contact 150 businesses, it is reasonable to use the B(150; 0:5) distribution for the number X who respond. Explain why. (b) What is the expected number (the mean) who will respond? (c) What is the probability that 70 or fewer will respond? (d) How large a sample must you take to increase the mean number of respondents to 100? (a) There are 150 independent observations, each with probability of success (a response) equal to 0.5. (b) = np = (150)(0.5) = 75. (c) P(X 70) = (exact result using software), or (using normal approx. with continuity correction, and normal table in text: z = 0.73), or (using normal approx. without continuity correction, and normal table in text: z = 0.82). (d) Use n = 200, since (200)(0.5) =

13 S5.6 According to government data, 21% of American children under the age of six live in households with incomes less than the official poverty level. A study of learning in early childhood chooses an SRS of 300 children. (a) What is the mean number of children in the sample who come from poverty-level households? What is the standard deviation of this number? (b) Use the normal approximation to calculate the probability that at least 80 of the children in the sample live in poverty. Be sure to check that you can safely use the approximation. (a) = (300)(0.21) = 63, = ( 300)(0.21)(0.79) = (b) np = 63 and n(1 p) = 237 are both more than 10, so we may approximate using the normal distribution: P(X 80) = P(Z 2.41) = , or with the continuity correction: P(X 79.5) = P(Z 2.34) =

14 Question State whether the following statements are true or false. (i) As the sample size increases, the mean of the sampling distribution of the sample mean ( X ) decreases. (ii) As the sample size increases, the standard deviation of the sampling distribution of the sample mean ( X ) decreases. (iii) The mean ( X ) of a random sample of size 4 is from a negatively skewed distribution is approximately normally distributed. (iv) The distribution of the proportion of successes ( ˆp ) in a sufficiently large sample is approximately normal with mean p and standard deviation np(1 p) where p is the population proportion and n is the sample size. (v) If X is the mean of a simple random sample of size 9 from N(500, 18) distribution, then X has a normal distribution with mean 500 and variance

A random variable (r. v.) is a variable whose value is a numerical outcome of a random phenomenon.

Chapter 14: random variables p394 A random variable (r. v.) is a variable whose value is a numerical outcome of a random phenomenon. Consider the experiment of tossing a coin. Define a random variable