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RoxasAngel
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Tyro
Asked: 2021-04-11T14:36:57+00:00 In: ,

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The average salary of a sales lady in malls in Metro Manila is P 430/ a day. Assume a normal distribution with a standard deviation of P 80.What is the probability that a random selected sale lady is receiving a daily salary bigger than 500?

b. For a sample of 30 sale lady, what is the probability that the sale lady’s salary is smaller than P 400 in a day?

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  2. acedstud

    a.) Let X be a random variable representing the salary of a saleslady in malls in Metro Manila, then the probability that X is less than a value x is given by P(X<x)=P\left(z<\dfrac{x-\mu}{\sigma}\right), where \mu is the population mean and \sigma is the population standard deviation.

    Here, \mu=430, \sigma=80.

    Thus, the probability that a randomly selected sale lady is receiving a daily salary bigger than 500 is given by P(X>500)=1-P(X<500)=1-P\left(z<\dfrac{500-430}{80}\right)=1-P(z<0.875)

    Using the table of areas under the standard normal curve or a calculator, we have =1-0.80921=0.19079.

    Therefore, the probability that a randomly selected sales lady is receiving a daily salary bigger than 500 is 0.19079.

     

    b.) Let \bar{Y} be a random variable representing the average salary of a saleslady in malls in Metro Manila from a sample of size n, then the probability that \bar{Y} is less than a value \bar{y} is given by P(\bar{Y}<\bar{y})=P\left(z<\dfrac{\bar{y}-\mu}{\sigma/\sqrt{n}}\right), where \mu is the population mean and \sigma is the population standard deviation.

    Here, \mu=430, \sigma=80, and n=30.

    a.) Thus, the probability that the saleladys’ average salary is smaller than P 400 in a day is given by P(\bar{Y}<400)=P\left(z<\dfrac{400-430}{80/\sqrt{30}}\right)=P(z<-2.054)

    Using the table of areas under the standard normal curve or a calculator, we have =0.01999.

    Therefore, the probability that the saleladys’ average salary is smaller than P 400 in a day is 0.01999.

     

    b.) The probability that the mean of private vehicles ages is over 6 years is given by P(\bar{X}>1)=1-P(\bar{X}<6)=1-P\left(z<\dfrac{6-6}{2/\sqrt{10}}\right)=1-P(z<0)

    Using the table of areas under the standard normal curve or a calculator, we have =1-0.5=0.5.

    Therefore, the probability that the mean of private vehicles ages is over 6 years is 0.5.

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