Chapter 1. The Normal Approximation to the Binomial Distribution

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.

2

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Correct.In this case, np = 60*0.86 = 51.6. This is larger than 10, so we check n(1 - p) = 60(1 – 0.86) = 60*0.14 = 8.4. This is not larger than 10, so we cannot use a Normal approximation.

Incorrect. In this case, np = 60*0.86 = 51.6. This is larger than 10, so we check n(1 - p) = 60(1 – 0.86) = 60*0.14 = 8.4. This is not larger than 10, so we cannot use a Normal approximation.

2

Try again.

Correct. In this case, np = 55*0.2 = 11. This is larger than 10, so we check n(1 - p) = 55(1 – 0.2) = 55*0.8 = 44. This is also larger than 10, so we can use a Normal approximation.

Incorrect. In this case, np = 55*0.2 = 11. This is larger than 10, so we check n(1 - p) = 55(1 – 0.2) = 55*0.8 = 44. This is also larger than 10, so we can use a Normal approximation.

2

Try again.

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.

2

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Correct. In this case, we have a small value of p, but a large n. Here, we will expect a bell-shaped, symmetric distribution.

Incorrect. In this case, we have a small value of p, but a large n. Here, we will expect a bell-shaped, symmetric distribution.

2

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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 \(\sigma = \sqrt{100*0.23*(1-0.23)}\) = 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 \(\sigma = \sqrt{100*0.23*(1-0.23)}\) = 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.

2

Try again.

Correct. This is a large sample with p close to 0.5. The mean is µ = 1036*0.54 = 559.44 (much larger than 10). Similarly, we can see that n(1 - p) will also be much larger than 10. The standard deviation is \(\sigma = \sqrt{1036*0.54*(1-0.54)}\) = 16.042. The z-score for X = 525 is z = (525 – 559.44)/16.042 = -2.15, rounded to two decimal places. Using the standard Normal table, we find the probability is 0.0158, or 0.016 when rounded to three decimal places.

Incorrect. This is a large sample with p close to 0.5. The mean is µ = 1036*0.54 = 559.44 (much larger than 10). Similarly, we can see that n(1 - p) will also be much larger than 10. The standard deviation is \(\sigma = \sqrt{1036*0.54*(1-0.54)}\) = 16.042. The z-score for X = 525 is z = (525 – 559.44)/16.042 = -2.15, rounded to two decimal places. Using the standard Normal table, we find the probability is 0.0158, or 0.016 when rounded to three decimal places.

2

Try again.