Answer to Question #130710 in Statistics and Probability for Enny

Question #130710
2. The annual salaries of employees in a large company are approximately normally distributed with a mean of N50,000 and a standard deviation of N 20,000.
i. What percent of people earn less than N 40,000?
ii. What percent of people earn between N 45,000 and N 65,000?
iii. What percent of people earn more than N 70,000?
1
Expert's answer
2020-08-26T17:15:03-0400

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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