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Continued: Normal Approximation for a Binomial Distribution Function Example 3: MathDrink, a soft-drink company, knows that it has a 42% market share in one region of the province. MathDrink's marketing department conducts a blind taste test of 70 people at the local mall. a) What is the probability that fewer than 25 people will choose MathDrink? b) What is the probability that exactly 25 people will choose MathDrink? Solution: Let X be the random variable representing the number of people that choose MathDrink. a) We have n=70, with p=0.42 and q=0.58. Since nxp=70x0.42=29.4>5 and nq=70x0.58=40.6>5 then we can use the normal approximation. mean=29.4 and S.D.=4.13 So, P(X<25) can be approximated to P(X<=24.5) (Recall Continuity Correction from last class). P(X<24.5)=P(Z<(24.5-29.4)/4.13)=P(Z<-1.19)=0.117 So, there is approx. 12% probability that fewer than 25 people surveyed will choose MathDrink. b) P(X=10) can be approximated to P(24.5<X<25.5). P(24.5<X<25.5)=P(-1.19<Z<-0.94)=P(Z<-0.94)-P(Z<-1.19)=0.057 The probability that exactly 25 people will choose MathDrink is approximately 5.7%. Example 4: A multiple choice exam contains 50 questions and each question has four answers from which to choose. If a student merely guesses at the answers, what is the probability that a) the student will get ten questions correct? b) the student will pass? Solution: Let X be the random variable representing the number of correct answers. We have n=50 Benroulli trials with p=1/4=0.25 and q=1-p=0.75. a) Since nxp=50x0.25=12.5>5 and nxq=50x0.0.75=37.5>5 then we can use the normal approximation. mean=nxp=12.5 and S.D.=squareRoot(nxpxq)=3.06. So, P(X=10) in the Binomial Distribution can be approximated to P(9.5<X<10.5) in the normal distribution. P(9.5<X<10.5)= P(X<10.5) – P(X<9.5) = P(Z<-0.65)-P(Z<-0.98) = 0.2578-0.1635 = 0.0943 = 9.4% Therefore, there is a probability of approximately 9.4% that the student gets 10 questions correct by guessing. b) The student needs to get at least 25 questions correct in order to pass the exam. So, P(X=>25) can be approximated to P(X>24.5). P(X>24.5)=1-P(X<24.5) =1- P(Z<3.92) =1-1 =0 Therefore, the probability that the student will pass the exam is 0%.