# Chapter 1. Binomial Mean and Standard Deviation

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2:04

### Question 1.1

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Correct. The mean (expected value) is μ = np = 25*0.23.
Incorrect. The mean (expected value) is μ = np = 25*0.23.
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### Question 1.2

EZ/LQ+Bufc3H3AtlDq9UfxFzjC6QXyh9JbxSTbepuH1TEPDESuTCj7i8GgT2ETbGrXD3QsVjz2OJ4nfZEROG3V9d+Ij1XD5QG2xi4PWPSRAS4Bo/OBGiIubz2mfkGBZfNBdagZzURl9tH9+uJJ3Rd4G62lLKJ8HrDsQdULP0y3H/nhPexMR4tEZLO3fEILDrYtNVSWAqfQZ2hswmB1ge0PI6qS28wOti5v+fLyQjYI7BoxA/+fdrSXxPtIZcd555MtIO5CH3WC4d4HTOhMVB+eoQA83eBGX+S6Hipm67xKwsdpQk
Correct.The standard deviation is $$\sigma = \sqrt{25*0.23*(1-0.23)} = \sqrt{4.4275}$$.
Incorrect. The standard deviation is $$\sigma = \sqrt{25*0.23*(1-0.23)} = \sqrt{4.4275}$$.
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### Question 1.3

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Correct.The mean (expected value) is μ = np = 1036*0.54.
Incorrect. The mean (expected value) is μ = np = 1036*0.54.
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### Question 1.4

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Correct. The standard deviation is $$\sigma = \sqrt{1036*0.54*(1-0.54)} = \sqrt{257.3424}$$.
Incorrect. The standard deviation is $$\sigma = \sqrt{1036*0.54*(1-0.54)} = \sqrt{257.3424}$$.
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5:56

### Question 1.5

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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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### Question 1.6

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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.
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