he mean number of words per minute (WPM) read by sixth graders is 95 with a variance of 196 . If 140 sixth graders are randomly selected, what is the probability that the sample mean would be less than 92.72 WPM? Round your answer to four decimal places.

Answers

Answer 1

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 probability

We 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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