Let "Z=\\frac{t-\\mu}{\\sigma}" where "t" is period of time "\\mu" is mean value of t and "\\sigma" is standard deviation of t.
Z has standard normal distribution
"P(a<t\\leq b)=P(\\frac{a-\\mu}{\\sigma}<\\frac{t-\\mu}{\\sigma}\\leq\\frac{b-\\mu}{\\sigma})=\\\\\nP(\\frac{a-\\mu}{\\sigma}<Z\\leq\\frac{b-\\mu}{\\sigma})=P(\\frac{a-25}{4}<Z\\leq\\frac{b-25}{4})="
"P(Z\\leq\\frac{b-25}{4})-P(Z\\leq\\frac{a-25}{4})"
1)
"P(t>30)=1-P(Z\\leq\\frac{30-25}{4})=\\\\\n1-P(Z\\leq\\frac{5}{4})=1-0.8944=0.1056"
2)
"P(18<t<34)=\\\\\nP(Z\\leq \\frac{34-25}{4}=2.25)-P(Z\\leq \\frac{18-25}{4}=-1.75)=\\\\\n0.9878-0.0401=0.9477"
3)
"P(25<t<34)=\\\\\nP(Z\\leq \\frac{34-25}{4}=2.25)-P(Z\\leq \\frac{25-25}{4}=0)=\\\\\n0.9878-0.5=0.4878"
Answer: "P(t>30)=0.1056" , "P(18<t<34)=0.9477" , "P(25<t<34)=0.4878"
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