SECTION 4.2 Binomial Distributions 209 Mean, Variance, and Standard Deviation Although you can use the formulas you learned in Section 4.1 for mean, variance, and standard deviation of a discrete probability distribution, the properties of a binomial distribution enable you to use much simpler formulas. Mean: m = np Variance: s 2 = npq Standard deviation: s = 2npq Population Parameters of a Binomial Distribution Finding and Interpreting Mean, Variance, and Standard Deviation In Pittsburgh, Pennsylvania, about 56% of the days in a year are cloudy. Find the mean, variance, and standard deviation for the number of cloudy days during the month of June. Interpret the results and determine any unusual values. (Source: National Oceanic and Atmospheric Administration) SOLUTION There are 30 days in June. Using n = 30, p = 0.56, and q = 0.44, you can find the mean, variance, and standard deviation as shown below. m = np = 30# 0.56 = 16.8 Mean s 2 = npq = 30# 0.56# 0.44 ≈ 7.4 Variance s = 2npq = 230# 0.56# 0.44 ≈ 2.7 Standard deviation Interpretation On average, there are 16.8 cloudy days during the month of June. The standard deviation is about 2.7 days. Values that are more than two standard deviations from the mean are considered unusual. Because 16.8 - 212.72 = 11.4 a June with 11 cloudy days or less would be unusual. Similarly, because 16.8 + 212.72 = 22.2 a June with 23 cloudy days or more would also be unusual. TRY IT YOURSELF 8 In San Francisco, California, about 44% of the days in a year are clear. Find the mean, variance, and standard deviation for the number of clear days during the month of May. Interpret the results and determine any unusual events. (Source: National Oceanic and Atmospheric Administration) Answer: Page A39 EXAMPLE 8
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