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a normal population has a mean of 59.8 and a standard deviation of 18.0. a sample of 10 will be taken. find the probability that the sample mean will be greater than 59.8. a) calculate the z score. (round your answer to 2 decimals.) b) find the probability that the sample mean will be greater than 59.8. (round your answer to 4 decimals.)

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MrRoyal

The z-score is 0 and the probability that the sample mean will be greater than 59.8 is 0.5000

How to determine the probability?

From the question, the given parameters about the distribution are

  • Mean value of the set of data = 59.8
  • Standard deviation value of the set of data = 18.0
  • The actual data value = 59.8

The z-score

The z-score of the data value is calculated using the following formula

z = (x - mean value)/standard deviation

Substitute the given parameters in the above equation

z = (59.8 - 59.8)/18.0

Evaluate

z = 0

The probability

The probability that the sample mean will be greater than 59.8 is then calculated as:

P(x > 59.8) = P(z > 0)

From the z table of probabilities, we have;

P(x > 59.8) = 0.5

Hence, the probability is 0.5000

Read more about probability at:

brainly.com/question/25870256

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