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....
View
Full Document
 Spring '14
 Statistics, grade point score, sample proportion

Click to edit the document details