g The average teachers salary in north Dakota is $37,764. Assume a normal distribution with LaTeX: \sigma=\text{5100}σ = 5100. a) For a samp

Question

g The average teachers salary in north Dakota is $37,764. Assume a normal distribution with LaTeX: \sigma=\text{5100}σ = 5100. a) For a sample of 75 teachers, what is the probability that the mean is greater than $38,000? Present your answer in 4 decimal places. b) What is the probability that a randomly selected teacher’s salary is grater than $45,000? Present your answer in three decimal places. Present your answers as: a:___,b:___

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Delilah 11 hours 2021-09-15T06:05:56+00:00 1 Answer 0

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    2021-09-15T06:07:29+00:00

    Answer:

    (a) Probability that the mean is greater than $38,000 is 0.3446.

    (b) Probability that a randomly selected teacher’s salary is grater than $45,000 is 0.078.

    Step-by-step explanation:

    We are given that the average teachers salary in north Dakota is $37,764. Assume a normal distribution with sigma (σ) = 5100.

    (a) A sample of 75 teachers is taken.

    Let \bar X = sample mean salary

    The z score probability distribution for sample mean is given by;

                                 Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

    where, \mu = population mean salary = $37,764

                \sigma = standard deviation = $5,100

                n = sample of teachers = 75

    Now, probability that the mean is greater than $38,000 is given by = P(\bar X > $38,000)

        P(\bar X > $38,000) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{38,000-37,764}{\frac{5,100}{\sqrt{75} } } ) = P(Z > 0.40) = 1 – P(Z < 0.40)

                                                                          = 1 – 0.6554 = 0.3446

    The above probability is calculated by looking at the value of x = 0.40 in the z table which has an area of 0.6554.

    (b) Let X = a randomly selected teacher’s salary

    The z score probability distribution for normal distribution is given by;

                                 Z  =  \frac{ X-\mu}{\sigma}  ~ N(0,1)

    where, \mu = population mean salary = $37,764

                \sigma = standard deviation = $5,100

    Now, probability that a randomly selected teacher’s salary is grater than $45,000 is given by = P(X > $45,000)

        P(X > $45,000) = P( \frac{ X-\mu}{\sigma} > \frac{45,000-37,764}{5,100} } ) = P(Z > 1.42) = 1 – P(Z < 1.42)

                                                                         = 1 – 0.9222 = 0.078

    The above probability is calculated by looking at the value of x = 1.42 in the z table which has an area of 0.9222.

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45:7+7-4:2-5:5*4+35:2 =? ( )