Let X will be the random variable denotes the salaries of the employees.
Mean μ = 16000
Standard deviation σ = 800
X ~ N(μ = 16000, σ = 800)
"Z = \\frac{X \u2013 \u03bc}{\u03c3} = \\frac{X \u2013 16000}{800} \\\\\n\n(a) \\; P(X > 18000) = P(\\frac{X \u2013 \u03bc}{\u03c3} > \\frac{18000 \u2013 16000}{800} )\\\\\n\n= P(Z > 2.5) \\\\\n\n= 1 \u2013 P(Z <= 2.5) \\\\\n\n= 1 \u2013 0.9937 \\\\\n\n= 0.0062 \\\\\n\n(b) \\; P(X < t) = 0.80 \\\\\n\nP(\\frac{X \u2013 \u03bc}{\u03c3} < \\frac{t \u2013 16000}{800}) = 0.80 \\\\\n\nP(Z < \\frac{t \u2013 16000}{800}) = 0.80 \\\\\n\n\\frac{t \u2013 16000}{800} = 0.8416 \\\\\n\nt = 16000 + 800 \\times 0.8416 \\\\\n\nt = 16673.28"
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