Given that,
Population mean "\\mu" = 50000
Standard deviation "\\sigma" = 20000
ANSWER 1)
We have to find the probability for total percentage of people 'Z' earning less than N 40000. which is given by Probability P(x<40000)
On converting the given data to a standard normal distribution
Z = "\\frac {x - \\mu}{\\sigma} \\sim N(0,1)"
Hence P(x<40000) = P("\\frac {x-\\mu}{\\sigma} < \\frac {40000 - \\mu}{\\sigma}" ) = P(Z < "\\frac {40000-50000}{20000}" ) = P(Z < -0.5)
From the standard normal table we get the value of
P(Z < -0.5) = 0.3085
ANSWER 2)
We have to find the probability for total percentage of people 'Z' earning between N 45000 to N 65000. which is given by Probability P(65000 < x < 45000)
On converting the given data to a standard normal distribution
Z = "\\frac {x - \\mu}{\\sigma} \\sim N(0,1)"
Hence,
P(65000 > x > 45000) = P("\\frac {65000 - \\mu}{\\sigma} >" "\\frac {x-\\mu}{\\sigma} > \\frac {45000 - \\mu}{\\sigma}" )
=("\\frac {65000 - 50000}{20000} >""\\frac {x-\\mu}{\\sigma} > \\frac {45000 - 50000}{20000}" )
P(65000 > x > 45000) = P(0.75 > Z > -0.25)
From the standard normal table we get the value of
P(0.75 > Z > -0.25) = (0.7734 - 0.5) + (0.5 - 0.4013) = 0.3721
ANSWER 3)
We have to find the probability for total percentage of people 'Z' earning more than N 70000. which is given by Probability P(x>70000)
On converting the given data to a standard normal distribution
Z = "\\frac {x - \\mu}{\\sigma} \\sim N(0,1)"
Hence P(x>70000) = P("\\frac {x-\\mu}{\\sigma} > \\frac {70000 - \\mu}{\\sigma}" ) = P(Z > "\\frac {70000-50000}{20000}" ) = P(Z > 1)
From the standard normal table we get the value of
P(Z > 1) = 0.5 - 0.3413 = 0.1587
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