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A recent study was conducted to investigate the duration of time required to complete a certain manual dexterity task. the reported mean was 10.2 seconds with a standard deviation of 16.0 seconds. suppose the reported values are the true mean and standard deviation for the population of subjects in the study. if a random sample of 144 subjects is selected from the population, what is the approximate probability that the mean of the sample will be more than 11.0 seconds?

Respuesta :

carlosego
It can be considered that the distribution of the sample is normal, based on the central limit theorem, where the sigma of the sample is:σ / root (n).
 The mean of the sample is:
 μm = μ
 So:
 P (X> 11) = P (Z> Zo).
 Where Z follows a standard normal distribution and
 Zo = (μm-μ) / (σ / root (n))
 Zo = (11-10.2) / (16 / root (144)).
 Zo = 0.6
 From the table for the standard normal distribution we have to:
 P (Z> 0.6) = 0.2743
fichoh

Using the Zscore relation, the approximate probability of having a sample mean greater than 11.0 seconds is 0.2739

The standard error = σ/√n = (16 /√144)

The standard error, S. E = 16/12 = 1.33

The Zscore :

  • Zscore = (x - μ) ÷ S.E

The Zscore = (11.0 - 10.2) ÷ 1.33

The Zscore = 0.601

The probability that mean will be greater than 11.0 seconds can be calculated thus :

P(Z > 0.601) = 1 - P(Z < 0.601)

Using a normal distribution table ;

P(Z < 0.601) = 0.72608

P(Z > 0.601) = 1 - 0.72608 = 0.2739

Therefore, the approximate probability of the sample mean being greater than 11.0 seconds is 0.2739.

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