StatTutor Lesson - The Normal Approximation to the Binomial Distribution

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Normal approx to binomial dist
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      Questions 1-3

      200

      Question 1.

      If n = 50 and p = 0.10, is n “large” enough to use a Normal approximation to the binomial distribution?

      A.
      B.

      Correct. In this case, np = 50*0.10 = 5. This is not larger than 10, so we can stop here.
      Incorrect. In this case, np = 50*0.10 = 5. This is not larger than 10, so we can stop here.
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      Questions 4-5

      364

      Question 4.

      If we have n = 25 and p = 0.7, we will expect the shape of the distribution to be

      A.
      B.
      C.

      Correct. Because p is larger than 0.5 and n is fairly small, we will expect a somewhat left-skewed shape.
      Incorrect. Because p is larger than 0.5 and n is fairly small, we will expect a somewhat left-skewed shape.
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      Questions 6-7

      511

      Question 6.

      At a large university, 23% of students are from out-of-state. If a random sample of 100 students is taken, what is the probability more than 30 are from out-of-state?

      A.
      B.
      C.
      D.

      Correct. In a sample of 100, we have mean µ = 100*0.23 = 23 (which is larger than 10) and n(1 - p) = 100*0.77 = 77 which is also larger than 10. We continue to find the standard deviation σ=1000.23(10.23) = 4.208. The z-score corresponding to X = 30 is z = (30 – 23)/4.208 = 1.66. Using the standard Normal table, we find the desired probability is 1 - P(z < 1.66) = 1 – 0.9515 = 0.0485.
      Incorrect. In a sample of 100, we have mean µ = 100*0.23 = 23 (which is larger than 10) and n(1 - p) = 100*0.77 = 77 which is also larger than 10. We continue to find the standard deviation σ=1000.23(10.23) = 4.208. The z-score corresponding to X = 30 is z = (30 – 23)/4.208 = 1.66. Using the standard Normal table, we find the desired probability is 1 - P(z < 1.66) = 1 – 0.9515 = 0.0485.
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