Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for grad

Question

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 181000 dollars. Assume the standard deviation is 31000 dollars. Suppose you take a simple random sample of 60 graduates.

Find the probability that a single randomly selected policy has a mean value between 172595.6 and 196608.1 dollars.
P(172595.6 < X < 196608.1) =
(Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a random sample of size
n
=
60
has a mean value between 172595.6 and 196608.1 dollars.
P(172595.6 < M < 196608.1) =
(Enter your answers as numbers accurate to 4 decimal places.)

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Khánh Gia 5 months 2021-09-01T10:48:50+00:00 1 Answers 3 views 0

Answers ( )

    0
    2021-09-01T10:50:04+00:00

    Answer:

    0.2979 = 29.79  probability that a single randomly selected policy has a mean value between 172595.6 and 196608.1 dollars.

    0.982 = 98.2% probability that a random sample of size 60 has a mean value between 172595.6 and 196608.1 dollars.

    Step-by-step explanation:

    To solve this question, we need to understand the normal probability distribution and the central limit theorem.

    Normal Probability Distribution:

    Problems of normal distributions can be solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

    Central Limit Theorem

    The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

    For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

    On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 181000 dollars. Assume the standard deviation is 31000 dollars.

    This means that \mu = 181000, \sigma = 31000

    Find the probability that a single randomly selected policy has a mean value between 172595.6 and 196608.1 dollars.

    This is the p-value of Z when X = 196608.1 subtracted by the p-value of Z when X = 172595.6. So

    X = 196608.1

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{196608.1 - 181000}{31000}

    Z = 0.5

    Z = 0.5 has a p-value of 0.6915

    X = 172595.6

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{172595.6 - 181000}{31000}

    Z = -0.27

    Z = -0.27 has a p-value of 0.3936

    0.6915 – 0.3936 = 0.2979

    0.2979 = 29.79  probability that a single randomly selected policy has a mean value between 172595.6 and 196608.1 dollars.

    Sample of 60:

    This means that n = 60, s = \frac{31000}{\sqrt{60}}

    Now, the probability is given by:

    X = 196608.1

    Z = \frac{X - \mu}{\sigma}

    By the Central Limit Theorem

    Z = \frac{X - \mu}{s}

    Z = \frac{196608.1 - 181000}{\frac{31000}{\sqrt{60}}}

    Z = 3.9

    Z = 3.9 has a p-value of 0.9999

    X = 172595.6

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{172595.6 - 181000}{\frac{31000}{\sqrt{60}}}

    Z = -2.1

    Z = -2.1 has a p-value of 0.0179

    0.9999 – 0.0179 = 0.982

    0.982 = 98.2% probability that a random sample of size 60 has a mean value between 172595.6 and 196608.1 dollars.

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