a) Using Excel we find mean, median, standard deviation. Using PERCENTILE.INC () we find the bottom 20% and top 10% performance level.
b) We build a histogram. From the histogram we can see that the distributiion is left-skewed (left tail is longer). Skewness equals -0.95. So the distribution is negatively skewed (mean<median). For normal distribution skewness is equal to 0. Kurtosis (whether the data are heavy-tailed or light-tailed) equals -0.07. We see that kurtosis does not differ much from 0. For standard normal distribution skewness is equal to 0. The distribution of the examination result does not look like a normal distribution.
c) "n=18, N=22, P=\\frac{18}{22}\\text{ is population proportion}."
"\\mu_{\\overline{p}}=P=\\frac{18}{22}.\\\\\n\\frac{n}{N}>0.05.\\text{ So we have: }\\\\\n\\sigma_{\\overline{p}}=\\sqrt{\\frac{P(1-P)}{n}}\\sqrt{\\frac{N-n}{N-1}}=\\frac{2}{11}\\sqrt{\\frac{1}{21}}.\\\\\nP\\{\\overline{p}>0.8\\}=1-P\\{\\overline{p}\\leq 0.8\\}=1-F(0.8)=\\\\\n=1-\\Phi(\\frac{0.8-\\frac{18}{22}}{\\frac{2}{11}\\sqrt{\\frac{1}{21}}})\\approx 1-0.3234=0.6766.\\\\"
Comments
Leave a comment