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8.4 Normal Approximation to the Binomial Distribution

8.4 Normal Approximation to the Binomial Distribution. Variables. n, the number of objects p, the probability of success q, the probability of failure. Conditions. In order to accurately approximate the binomial distribution… np > 5 nq > 5. Formulas.  =np σ= Z= n C x p x q n-x.

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8.4 Normal Approximation to the Binomial Distribution

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  1. 8.4 Normal Approximation to the Binomial Distribution

  2. Variables • n, the number of objects • p, the probability of success • q, the probability of failure

  3. Conditions • In order to accurately approximate the binomial distribution… • np > 5 • nq > 5

  4. Formulas • =np • σ= • Z= • nCxpxqn-x

  5. Ex. 1 It is estimated that 15% of the Canadian population would choose a pop/rock concert if they are given a choice of musical events A) If 35 Canadians are randomly surveyed, what is the probability that at least 5 of them will choose a pop/rock concert?

  6. Solution • Solution n=35 p=0.15 q=0.85 np=(35)(0.15) nq=(35)(0.85) =5.25>5 =29.75>5 P(X>5) = P(X>4.5) = 1- P(X<4.5) P(X>5) = 1 – 0.3632 P(X>5) = 0.6368 P(X>5) = 63.68% Therefore of the group of 30, at least 5 Canadians will choose a pop/rock concert 63.68%

  7. B) Compare the results from the normal approximation with the results from the calculation using a binomial distribution

  8. Solution Solution n= 35 p=0.15 q=0.85 We know from unit 7 that : P(X)= nCxpxqn-x We want probability of X ≥ 5, we can add all the cases where X =5, X=6, X=7 , …… till X=35 P(X=5)=35C5(0.15)5(0.85)30=0.188111521 P(X=6)=35C6(0.15)6(0.85)29=0.165980753 P(X=35)=35C35(0.15)35(0.85)0=1.456109606-29 0.619349782 Therefore the normal approximation gives us a probability of 63.68%, while the binomial distribution gives us an answer of 61.93%

  9. Ex. 2 A manufacturer of toys recalled one of its products when it was discovered that 30% of the toys had a defective part. In a shipment of 600 of these toys, what is the probability that fewer than 200 are defective?

  10. Solution n=600 p=0.3 q=0.7 np=(600)(0.3) nq=(600)(0.7) =(600)(0.3) σ= np=180>5 nq=420>5 =180 σ=11.22 P(X<199.5)=P(Z<1.74) =0.9591 =95.91% Therefore, the probability that less than 200 toys are defective is 95.91%.

  11. Ex. 3 It is estimated that 89% of Canadian women are 150 cm or taller. If a poll of 400 Canadian women is conducted, what is the probability that more than 360 of them are 150 or taller? • State whether a normal distribution approximation is reasonable for this binomial distribution situation. B) Determine the indicated probability.

  12. Solution n=400 p=0.89 q=0.11 np=(400)(0.89) np=(400)(0.11) np=356>5 np=44>5 Therefore a normal distribution approximation is reasonable.

  13. Solution =np =(400)(0.89) =356 σ= σ= σ=6.26 =1-0.7389 =0.2611 =26.11% Therefore, the probability that more than 360 Canadian women are taller than 150cm is 26.11%

  14. Pg 449 # 1-6,8-10

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