The probability that the sample mean would be less than 92.72 WPM is approximately 0.0107 or 1.07% (rounded to four decimal places).
The probabilityWe can use the Central Limit Theorem to approximate the distribution of the sample mean.
According to the problem, the population mean is μ = 95 and the variance is σ^2 = 196. Therefore, the population standard deviation is σ = sqrt(196) = 14.
The distribution of the sample mean can be approximated by a normal distribution with mean μx = μ = 95 and standard deviation σx = σ/sqrt(n) = 14/sqrt(140) ≈ 1.18.
To find the probability that the sample mean would be less than 92.72 WPM, we need to standardize the value of 92.72 using the formula:
z = (x - μx) / σx
where x is the value of the sample mean we are interested in, and μx and σx are the mean and standard deviation of the distribution of the sample mean.
Plugging in the values, we get:
z = (92.72 - 95) / 1.18 ≈ -2.33
Using a standard normal distribution table or calculator, we can find that the probability of obtaining a z-score less than -2.33 is approximately 0.0107.
Therefore, the probability that the sample mean would be less than 92.72 WPM is approximately 0.0107 or 1.07% (rounded to four decimal places).
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