Question

Suppose that Y has a binomial distribution with p = 0.40.

(a) Use technology and the normal approximation to the binomial distribution to compute the exact and approximate values of P(Y ≤ μ + 2) for n = 5, 10, 15, and 20. For each sample size, pay attention to the shapes of the binomial histograms and to how close the approximations are to the exact binomial probabilities. (Round your answers to five decimal places.)

n = 5

exact value P(Y ≤ μ + 2) =

approximate value P(Y ≤ μ + 2) ≈

n = 10

exact value P(Y ≤ μ + 2) =

approximate value P(Y ≤ μ + 2) ≈

n = 15

exact value P(Y ≤ μ + 2) =

approximate value P(Y ≤ μ + 2) ≈

n = 20

exact value P(Y ≤ μ + 2) =

approximate value P(Y ≤ μ + 2) ≈

b) Refer to part (a). What did you notice about the shapes of the binomial histograms as the sample size increased?

1.) The binomial histograms appear less mound shaped as the sample size increases.

2.) The binomial histograms appear the same as the sample size increases.

3.) The binomial histograms appear more mound shaped as the sample size increases.

(c) According to the rule of thumb for the adequacy of the normal approximation, how large must n be for the approximation to be adequate? (Round your answer up to the nearest whole number.)

n >

Is this consistent with what you observed in parts (a) and (b)?

1.) Yes, because the approximations for larger values of n are much better than the approximations for smaller values of n.

2.) No, because the approximations for smaller values of n are just as good as the approximations for larger values of n.

Answer 1

mean = np

sd = sqrt(npq)

P(Y <= mu +2)

= P(Y < mu + 2.5) continuity correction

a)

n	mean	μ + 2	exact	approximation
5	2	4	0.98976	0.98876
10	4	6	0.94524	0.94671
15	6	8	0.90495	0.90618
20	8	10	0.87248	0.87308

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b)
3.) The binomial histograms appear more mound shaped as the sample size increases.

c)
np > 10 , n(1-p) > 10

1.) Yes, because the approximations for larger values of n are much better than the approximations for smaller values of n.