# 60 the standard deviationp1pn0600405000048006928

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Unformatted text preview: d, x = 0 if a male is selected. *Population distribution: P(female) = P(1) = 0.60, P(male) = P(0) = 0.40 *Sample Data distribution: P(1) = 26/50 = 0.52, P(0) = 0.48 *Sampling distribution of the sample proportion of females, P(1) is an approximately normal distribution. The mean = P(1) = 0.60, the standard deviation=√[p(1‐p)/n]=√(0.60*0.40/50) = √0.0048 = 0.06928 *refer to next page for the graphs ST1131 Introduction to Statistics Tutorial 6 Population Distribution of X 0.6 0.5 P(X) 0.4 0.3 0.2 0.1 0.0 0 1 X Data Distribution of X 0.6 0.5 P(X) 0.4 0.3 0.2 0.1 0.0 0 1 X Sampling Distribution P(sample proportion) 6 5 4 3 2 1 0 0.4 0.5 0.6 0.7 0.8 Sample proportion of females 4 a. Population distribution: µ = \$74,550, σ = \$19,872 The shape is right‐skewed as a few employees receive high income. b. Data distribution: x‐bar = \$75,207, s = \$18,901 The shape is likely right‐skewed as it is a random sample. c. Sampling distribution of the sample mean, x‐bar for n = 100: µ = \$74,550, SE = \$19,872/√100 = \$1,987.20. The shape is approximately normal as the sample size is large. d. Income of \$100,000 is (100,000‐74,550)/19,872 = 1.28 standard deviations above the mean income in the population, thus it is not unusual. While a sample mean income of \$100,000 is (100,000‐ 74,550/1,987.20) = 12.8 standard errors above the mean in the sampling distribution, thus highly unusual....
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