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The mean number of words per minute (WPM) typed by a speed typist is 123123 with a standard deviation of 1414 WPM. What is the probability that the sample mean would be greater than 122.2122.2 WPM if 5252 speed typists are randomly selected

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Answer:

0.6598

Step-by-step explanation:

Given that:

Mean words per minute (m) = 123

Standard deviation (σ) = 14

Sample mean = 122.2

Sample size (n) = 52

Using the relation to obtain the Zscore :

Zscore = (sample mean - m) / (σ/√n)

Zscore = (122.2 - 123) / (14/√52)

Zscore = - 0.8 / 1.9415

Zscore = - 0.412

P(Z > - 0.412) = 1 - P(Z < - 0.412)

P(Z < - 0.412) = 0.34017

P(Z > - 0.412) = 1 - 0.34017

P(Z > - 0.412) = 0.6598

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