Day 46 Agenda: DG 20 --- 15 minutes

Advanced Placement Statistics Section 8.1 (Part II): Mean, Variance, and Standard Deviation for a Binomial Random Variable EQ: How do you calculate the mean, variance, and standard deviation for a binomial random variable? Get out the Ex 8.1 Worksheet from yesterday (President’s Physical Fitness Test).

Refer back to the Section 8 Refer back to the Section 8.1 Ex Handout on the President’s Physical Fitness Test to answer #1 - 3. Calculate the mean, variance and standard deviation, respectively, of this probability distribution. HINT: Create a probability distribution for the random variable X. (0)(.6561)+(1)(.2916)+(2)(.0486)+(3)(.0036)+(4)(.0001) = 0.4 REMEMBER: This is a COUNT, not a PERCENTAGE. 0.4 (0-.4)2(.6561)+(1-.4)2(.2916)+(2-.4)2(.0486)+(3-.4)2(.0036) +(4-.4)2(.0001) = 0.36

2. a. What are the values for the parameters?   n = ______ p = _______ q = _____ 4 0.1 0.9 (4)(.1) = 0.4 b. Find np. How does it compare to ? Find npq. (4)(.1)(.9) = 0.36

FORMULAS TO KNOW: For a binomial random variable, SEE FORMULA SHEET:

RECALL: NORMAL DISTRIBUTION .4 , .6 -1.4 -.8 -.2 .4 1.0 1.6 2.2 b. The Empirical Rule states approximately 95% of the x values lie within ­­­­____ standard deviations 2

This occurs from ___ to __ on your density curve. -.8 1.6 d. Since X cannot be negative, we would expect (in the long run), a large majority (95%) of the adults (out of groups of four) who pass the fitness test to be between _______ and _____. 1.6

# of successes # of failures Binomial Distribution Normal Approximation If: n is large enough implies n and p satisfy np > 10 and n(1 – p) > 10 # of successes # of failures Describes distribution mean Standard deviation

In Class Assignment: p. 529- 530 #22, 23 Get in groups so we can go over these problems together. Get out a sheet of notebook paper.

This distribution can be described as B(509,.116). p. 529 #22 X = number of homeruns out of 509 randomly selected at bats by Mark McGuire in 1998 Criteria for a Binomial Distribution: Fixed number of trials; n = 509 at bats Success  hit a homerun Failure  did not hit a homerun Result of one at bat is independent of the result of another at bat. Probability of hitting a homerun remains the same from trial to trial. p = .116 q = .884 This distribution can be described as B(509,.116).

REMEMBER: THIS IS A BINOMIAL DISTRIBUTION Find the expected number of homeruns we could expect Mark McGuire to have hit in 509 randomly selected at bats. REMEMBER: THIS IS A BINOMIAL DISTRIBUTION On average, in the long run, we would expect Mark McGuire to hit 59.044 homeruns in 509 at bats. Use a binomial distribution to calculate the probability that Mark McGuire will hit at least 70 homeruns in 509 at bats. The probability that Mark McGuire will hit at least 70 homeruns in 509 at bats is 7.64%. 1 - binomialcdf (509, 0.116, 69)

*Is Normal Approximation Appropriate? np > 10 ? nq > 10 ? (509)(.116) > 10 (509)(.884) > 10 59.044 > 10 449.96 > 10 Criteria met; Normal Approximation Appropriate RECALL: What does this allow us to do? Standardize using z scores

Standardize using z scores The probability that Mark McGuire will hit at least 70 homeruns in 509 at bats is 6.43%. How does the probability of Mark McGuire hitting at least 70 homeruns using a normal distribution compare to the probability calculated using a binomial distribution? The normal distribution gave a much smaller probability of 6.4% compared to 7.6% for the binomial distribution.

This distribution can be described as B(400,.92). p. 530 #23 X = number of people out of a random sample of 400 adults in Richmond who approve of the President’s response Criteria for a Binomial Distribution: Fixed number of trials; n = 400 people Success  approve President’s response Failure  did not approve President’s response One person’s opinion is independent of another person’s opinion. Probability of person approving President’s response remains the same from trial to trial. p = .92 q = .08 This distribution can be described as B(400,.92).

Use a binomial distribution to calculate the probability that at most 358 of the 400 adults in the Richmond poll approve of the President’s response. binomialcdf (400, 0.92, 358) The probability that at most 358 out of 400 adults in the Richmond poll approve of the President’s response to the WTC attacks is 4.41%.

REMEMBER: THIS IS A BINOMIAL DISTRIBUTION Calculate the expected number of adults in Richmond that approve of the President’s response. REMEMBER: THIS IS A BINOMIAL DISTRIBUTION On average, in the long run, we would expect 368 adults in Richmond to approve of the President’s response to the WTC attacks. Calculate the standard deviation for the number of adults in Richmond that approve of the President’s response.

*Is Normal Approximation Appropriate? np > 10 ? nq > 10 ? (400)(.92) > 10 (400)(.08) > 10 368 > 10 32 > 10 Criteria met; Normal Approximation Appropriate RECALL:

Standardize using z scores The probability that at most 358 adults in Richmond approve of the President’s response to the WTC attacks is 3.29%. The normal distribution gave a much smaller probability of 3.3% compared to 4.4% for the binomial distribution. How does the probability calculated using a normal distribution compare to the probability calculated using a binomial distribution?

ACCURACY OF NORMAL APPROXIMATION: Increases when p is close to ½. Less accurate when p is close to 0 or 1. RULE OF THUMB: BINOMIAL OR NORMAL ? If n, p, and q satisfy the conditions np > 10 and nq > 10 you are EXPECTED to use a Normal Approximation instead of a Binomial Distribution.

Assignment : WS Mean, Variance and Standard Deviation for Binomial Random Variable