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Sec. 7-6: The Normal Approximation to the Binomial Distribution

Sec. 7-6: The Normal Approximation to the Binomial Distribution. We’ve solved binomial problems using Table B—we now want to apply the concept of the normal distribution curve to solve binomial problems.

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Sec. 7-6: The Normal Approximation to the Binomial Distribution

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  1. Sec. 7-6: The Normal Approximation to the Binomial Distribution

  2. We’ve solved binomial problems using Table B—we now want to apply the concept of the normal distribution curve to solve binomial problems. • We can make something called a “continuity correction” to our binomial information to solve it like a normal distribution, as long as the following exists: np ≥ 5 AND nq ≥ 5

  3. Continuity Correction: Add or subtract .5 onto/from the x value. • To solve binomial P(x)’s: CHECK np & nq a. Determine the mean (μ) and the standard deviation (σ) using the old formulas: μ = np and σ = √npq b. Draw a picture if needed & CONTINUITY CORRECT AS NEEDED.

  4. c. Use the “original” z-score formula to determine what value to look up to get your probability from Table E. z = x – μ σ ** We are back to the original z-score because we are trying to find the probability from a ‘real life’ data value—not a mean.

  5. Here’s an example: Northwest Airlines recently reported that its on-time arrival rate is 72%. Find the probability that among 80 randomly selected flights: a.) FEWER than 70 arrive on time. b.) AT LEAST 70 arrive on time. c.) EXACTLY 70 arrive on time.

  6. First, identify your information: n = 80 p = .72 so… q = .28 *** CHECK np ≥ 5 and nq ≥ 5 np = 57.6 and nq = 22.4 so we can do this problem like a normal distribution. ** We need to use this info. to determine μ, & σ. μ = (80)(.72) = 57.6 σ = √80(.72)(.28) = 4.0

  7. Draw your picture: you need to C.C. to 69.5. z = 69.5 – 57.6 4 z = 2.975  2.98 Look up 2.98 as usual, & because the shading goes thru the mean so add .5. P(x < 70 on time) = .9986

  8. b) At least 70 arrive on time: (you will need to C.C. to 69.5) Z = 69.5 – 57.6 = 2.98 4 Look up 2.98 on the chart and SUBTRACT from .5 because the shading is a tail. P(x ≥ 70 on time) = .0014

  9. c) Exactly 70 arrive on time: We will have to C.C. to BOTH sides of 70 because we’re trying to create a single value. So we will have to do the z-score twice. Z = 69.5 – 57.6 z = 70.5 – 57.6 4 4 Z = 2.98 z = 3.23 P(x = 70 on time) = .4999 - .4986 = .0013

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