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STAT E100

STAT E100. Section Week 7 – Binomial Distribution. Review The exams are graded, see your TA for your midterm exam! If you did not do as well as you hoped, remember, the midterm is worth 20 % (undergraduates) or 15% (graduate students).

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STAT E100

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  1. STAT E100 Section Week 7 – Binomial Distribution

  2. Review • The exams are graded, see your TA for your midterm exam! • If you did not do as well as you hoped, remember, the midterm is worth 20% (undergraduates) or 15% (graduate students). • And homework is worth 30% (undergraduates) and 25% (graduate students). • Interested in forming a study group? Email: STATE100studygroup@gmail.com

  3. Key Concepts: The binomial distribution is characterized by 4 properties: 1) Fixed number (n) of observations, or `trials’. 2) The n trial are all independent of each other 3) Each trial has two possible outcomes: `success’ or `failure’. 4) The probability (denoted by p) of success at each trial is constant.

  4. Sample Question #1 Harvard College is composed of about 52% women. For all students currently enrolled in Stat 104, 172 of the 381 students are women. Use this information for the following problem. a) What is the expected number of women in a simple random sample of 381 students from Harvard College? What is the standard deviation? b) What is the probability of selecting exactly 172 women in a sample of 381 students from Harvard College? c) What is the probability of selecting exactly 172 women or fewer in a sample of 381 students from Harvard College? d) What is the probability of selecting at least 172 women or fewer in a sample of 381 students from Harvard College [use software]?

  5. Sample Question #1 Harvard College is composed of about 52% women. For all students currently enrolled in Stat 104, 172 of the 381 students are women. Use this information for the following problem. a) What is the expected number of women in a simple random sample of 381 students from Harvard College? What is the standard deviation? Expected number =n*p=0.52*381= 198.12 standard deviation=sqrt( np(1-p) )= 9.75 b) What is the probability of selecting exactly 172 women in a sample of 381 students from Harvard College? c) What is the probability of selecting exactly 172 women or fewer in a sample of 381 students from Harvard College? d) What is the probability of selecting at least 172 women or fewer in a sample of 381 students from Harvard College [use software]?

  6. Sample Question #1 Harvard College is composed of about 52% women. For all students currently enrolled in Stat 104, 172 of the 381 students are women. Use this information for the following problem. a) What is the expected number of women in a simple random sample of 381 students from Harvard College? What is the standard deviation? Expected number =n*p=0.52*381= 198.12 standard deviation=sqrt( np(1-p) )= 9.75 b) What is the probability of selecting exactly 172 women in a sample of 381 students from Harvard College? P(X=172) = = 0.0011 c) What is the probability of selecting exactly 172 women or fewer in a sample of 381 students from Harvard College? d) What is the probability of selecting at least 172 women or fewer in a sample of 381 students from Harvard College [use software]?

  7. Sample Question #1 Harvard College is composed of about 52% women. For all students currently enrolled in Stat 104, 172 of the 381 students are women. Use this information for the following problem. a) What is the expected number of women in a simple random sample of 381 students from Harvard College? What is the standard deviation? Expected number =n*p=0.52*381 = 198.12 standard deviation=sqrt( np(1-p) ) = 9.75 b) What is the probability of selecting exactly 172 women in a sample of 381 students from Harvard College? P(X=172) = = 0.0011 c) What is the probability of selecting exactly 172 women or fewer in a sample of 381 students from Harvard College? P(X≤172) : means we should sum the probabitlies of each possible amount of women up to 172 P(X=172)+P(X=171)+ P(X=170)+ P(X=169)… P(X=0) YIKES! Or we can use the normal approximation of the binomial distribution. X~Bin(n=381, p =0.52)  N(μx=198.12, σx=9.75) Z score = (172 – 198.12)/ 9.75 = -2.68  from the z-table, p = 0.0037 d) What is the probability of selecting at least 172 women or fewer in a sample of 381 students from Harvard College [use software]?

  8. Sample Question #1 Harvard College is composed of about 52% women. For all students currently enrolled in Stat 104, 172 of the 381 students are women. Use this information for the following problem. a) What is the expected number of women in a simple random sample of 381 students from Harvard College? What is the standard deviation? Expected number =n*p=381*0.52= 198.12 standard deviation=sqrt( np(1-p) )= 9.75 b) What is the probability of selecting exactly 172 women in a sample of 381 students from Harvard College? P(X=172) = = 0.0011 c) What is the probability of selecting exactly 172 women or fewer in a sample of 381 students from Harvard College? P(X≤172) : means we should sum the probabitlies of each possible amount of women up to 172 P(X=172)+ P(X=171)+ P(X=170)+ P(X=169)… P(X=0) YIKES! Or we can use the normal approximation of the binomial distribution. X~Bin(n=381, p =0.52)  N(μx=198.12, σx=9.75) Z score = (172 – 198.12)/ 9.75 = -2.68  from the z-table, p = 0.0037 d) What is the probability of selecting at least 172 women or fewer in a sample of 381 students from Harvard College [use software]? P(X≤172) = 0.0032 http://www.stat.tamu.edu/~west/applets/binomialdemo.html

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