Mathematics, 30.06.2021 23:30 claudiapineda860

The mean number of words per minute (WPM) read by sixth graders is 84 with a standard deviation of 15 WPM. If 188 sixth graders are randomly selected, what is the probability that the sample mean would differ from the population mean by greater than 1.46 WPM

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Answer from: QueensQueens

0.1836 = 18.36% probability that the sample mean would differ from the population mean by greater than 1.46 WPM

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean number of words per minute (WPM) read by sixth graders is 84 with a standard deviation of 15 WPM.

This means that $\mu = 84, \sigma = 15$

Sample of 188

This means that $n = 188, s = \frac{15}{\sqrt{188}}$

What is the probability that the sample mean would differ from the population mean by greater than 1.46 WPM?

Greater than 84 + 1.46 = 85.46 or less than 84 - 1.46 = 82.54. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Probability is is less than 82.54.

P-value of Z when X = 82.54. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{82.54 - 84}{\frac{15}{\sqrt{188}}}$

Z = -1.33

Z = -1.33 has a p-value of 0.0918

2*0.0918 = 0.1836

0.1836 = 18.36% probability that the sample mean would differ from the population mean by greater than 1.46 WPM

Answer from: Quest

yes

yes yes yes

Answer from: Quest

The answer to your question is b’(16,14)

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