The average time it takes a group of senior high school students to
complete a certain examination is 46.2 minutes. The standard deviation is
8 minutes. Assume that the variable is normally distributed.
What is the probability that a random selected college student will
complete the examination in less than 43 minutes?
Suppose that "X" is a random variable that has a normal distribution with parameters "\\mu=46.2" and "\\sigma=8". The respective probability density function is: "p(x)=\\frac{1}{\\sigma\\sqrt{2\\pi}}e^{-\\frac12\\left(\\frac{x-\\mu}{\\sigma}\\right)^2}". The aim is to find "P(X\\leq43)". We receive: "P(X\\leq43)=\\int_{-\\infty}^{43}\\frac{1}{\\sigma\\sqrt{2\\pi}}e^{-\\frac12\\left(\\frac{x-\\mu}{\\sigma}\\right)^2}dx\\approx0.3446" (it is rounded to "4" decimal places)
Answer: "0.3446" (it is rounded to "4" decimal places)
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